Binomial Coefficient | DP-9

The following are the common definitions of Binomial Coefficients. 

A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.

A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set.

The Problem
Write a function that takes two parameters n and k and returns the value of Binomial Coefficient C(n, k). For example, your function should return 6 for n = 4 and k = 2, and it should return 10 for n = 5 and k = 2.

1) Optimal Substructure
The value of C(n, k) can be recursively calculated using the following standard formula for Binomial Coefficients.  

   C(n, k) = C(n-1, k-1) + C(n-1, k)
   C(n, 0) = C(n, n) = 1

Following is a simple recursive implementation that simply follows the recursive structure mentioned above.  

C++

// A naive recursive C++ implementation
#include <bits/stdc++.h>
using namespace std;

// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    // Base Cases
    if (k > n)
        return 0;
    if (k == 0 || k == n)
        return 1;

    // Recur
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}

/* Driver code*/
int main()
{
    int n = 5, k = 2;
    cout << "Value of C(" << n << ", " << k << ") is "
         << binomialCoeff(n, k);
    return 0;
}

// This is code is contributed by rathbhupendra

C

// A Naive Recursive Implementation
#include <stdio.h>

// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    // Base Cases
    if (k > n)
        return 0;
    if (k == 0 || k == n)
        return 1;

    // Recur
    return binomialCoeff(n - 1, k - 1)
           + binomialCoeff(n - 1, k);
}

/* Driver program to test above function*/
int main()
{
    int n = 5, k = 2;
    printf("Value of C(%d, %d) is %d ", n, k,
           binomialCoeff(n, k));
    return 0;
}

Java

// JAVA Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
import java.util.*;

class GFG {

    // Returns value of Binomial
    // Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {

        // Base Cases
        if (k > n)
            return 0;
        if (k == 0 || k == n)
            return 1;

        // Recur
        return binomialCoeff(n - 1, k - 1)
            + binomialCoeff(n - 1, k);
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 5, k = 2;
        System.out.printf("Value of C(%d, %d) is %d ", n, k,
                          binomialCoeff(n, k));
    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# A naive recursive Python implementation


def binomialCoeff(n, k):

    if k > n:
        return 0
    if k == 0 or k == n:
        return 1

    # Recursive Call
    return binomialCoeff(n-1, k-1) + binomialCoeff(n-1, k)


# Driver Program to test ht above function
n = 5
k = 2
print ("Value of C(%d,%d) is (%d)" % (n, k,
                                     binomialCoeff(n, k)))

# This code is contributed by Nikhil Kumar Singh (nickzuck_007)

C#

// C# Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
using System;

class GFG {

    // Returns value of Binomial
    // Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {

        // Base Cases
        if (k > n)
            return 0;
        if (k == 0 || k == n)
            return 1;

        // Recur
        return binomialCoeff(n - 1, k - 1)
            + binomialCoeff(n - 1, k);
    }

    /* Driver program to test above function */
    public static void Main()
    {
        int n = 5, k = 2;
        Console.Write("Value of C(" + n + "," + k + ") is "
                      + binomialCoeff(n, k));
    }
}

// This code is contributed by Sam007.

Javascript

<script>
// javascript Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)

   // Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff(n , k)
{

    // Base Cases
    if (k > n)
        return 0;
    if (k == 0 || k == n)
        return 1;

    // Recur
    return binomialCoeff(n - 1, k - 1)
        + binomialCoeff(n - 1, k);
}

/* Driver program to test above function */
var n = 5, k = 2;
document.write("Value of C("+n+", "+k+") is "+binomialCoeff(n, k));

// This code is contributed by Amit Katiyar
</script>

PHP

<?php
// PHP Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)

// Returns value of
// Binomial Coefficient C(n, k)
function binomialCoeff($n, $k)
{
    // Base Cases
    if ($k > $n)
        return 0;
    if ($k==0 || $k==$n)
        return 1;

    // Recur
    return binomialCoeff($n - 1, $k - 1) +
               binomialCoeff($n - 1, $k);
}

    // Driver Code
    $n = 5;
    $k = 2;
    echo "Value of C","(",$n ,$k,") is "
               , binomialCoeff($n, $k);

// This code is contributed by aj_36
?>

Output

Value of C(5, 2) is 10

Time Complexity: O(n*max(k,n-k)) 

Auxiliary Space: O(n*max(k,n-k))

2) Overlapping Subproblems
It should be noted that the above function computes the same subproblems again and again. See the following recursion tree for n = 5 and k = 2. The function C(3, 1) is called two times. For large values of n, there will be many common subproblems. 

Binomial Coefficients Recursion tree for C(5,2)

Since the same subproblems are called again, this problem has the Overlapping Subproblems property. So the Binomial Coefficient problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, re-computations of the same subproblems can be avoided by constructing a temporary 2D-array C[][] in a bottom-up manner. Following is Dynamic Programming-based implementation. 

C++

// A Dynamic Programming based solution that uses
// table C[][] to calculate the Binomial Coefficient
#include <bits/stdc++.h>
using namespace std;

// Prototype of a utility function that
// returns minimum of two integers
int min(int a, int b);

// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;

    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= min(i, k); j++) {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;

            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }

    return C[n][k];
}

// A utility function to return
// minimum of two integers
int min(int a, int b) { return (a < b) ? a : b; }

// Driver Code
int main()
{
    int n = 5, k = 2;
    cout << "Value of C[" << n << "][" << k << "] is "
         << binomialCoeff(n, k);
}

// This code is contributed by Shivi_Aggarwal

C

// A Dynamic Programming based solution
// that uses table C[][] to
// calculate the Binomial Coefficient
#include <stdio.h>

// Prototype of a utility function that
// returns minimum of two integers
int min(int a, int b);

// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    int C[n + 1][k + 1];
    int i, j;

    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= min(i, k); j++) {
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;

            // Calculate value using
            // previously stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }

    return C[n][k];
}

// A utility function to return
// minimum of two integers
int min(int a, int b) { return (a < b) ? a : b; }

/* Driver program to test above function*/
int main()
{
    int n = 5, k = 2;
    printf("Value of C(%d, %d) is %d ", n, k,
           binomialCoeff(n, k));
    return 0;
}

Java

// A Dynamic Programming based
// solution that uses table C[][] to
// calculate the Binomial Coefficient

class BinomialCoefficient {
    // Returns value of Binomial
    // Coefficient C(n, k)
    static int binomialCoeff(int n, int k)
    {
        int C[][] = new int[n + 1][k + 1];
        int i, j;

        // Calculate  value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= min(i, k); j++) {
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;

                // Calculate value using
                // previously stored values
                else
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }

        return C[n][k];
    }

    // A utility function to return
    // minimum of two integers
    static int min(int a, int b) { return (a < b) ? a : b; }

    /* Driver program to test above function*/
    public static void main(String args[])
    {
        int n = 5, k = 2;
        System.out.println("Value of C(" + n + "," + k
                           + ") is " + binomialCoeff(n, k));
    }
}
/*This code is contributed by Rajat Mishra*/

Python3

# A Dynamic Programming based Python
# Program that uses table C[][]
# to calculate the Binomial Coefficient

# Returns value of Binomial Coefficient C(n, k)


def binomialCoef(n, k):
    C = [[0 for x in range(k+1)] for x in range(n+1)]

    # Calculate value of Binomial
    # Coefficient in bottom up manner
    for i in range(n+1):
        for j in range(min(i, k)+1):
            # Base Cases
            if j == 0 or j == i:
                C[i][j] = 1

            # Calculate value using
            # previously stored values
            else:
                C[i][j] = C[i-1][j-1] + C[i-1][j]

    return C[n][k]


# Driver program to test above function
n = 5
k = 2
print("Value of C[" + str(n) + "][" + str(k) + "] is "
      + str(binomialCoef(n, k)))

# This code is contributed by Bhavya Jain

C#

// A Dynamic Programming based solution that
// uses table C[][] to calculate the Binomial
// Coefficient
using System;

class GFG {

    // Returns value of Binomial Coefficient
    // C(n, k)
    static int binomialCoeff(int n, int k)
    {
        int[, ] C = new int[n + 1, k + 1];
        int i, j;

        // Calculate value of Binomial
        // Coefficient in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= Math.Min(i, k); j++) {
                // Base Cases
                if (j == 0 || j == i)
                    C[i, j] = 1;

                // Calculate value using previously
                // stored values
                else
                    C[i, j] = C[i - 1, j - 1] + C[i - 1, j];
            }
        }

        return C[n, k];
    }

    // A utility function to return minimum
    // of two integers
    static int min(int a, int b) { return (a < b) ? a : b; }

    /* Driver program to test above function*/
    public static void Main()
    {
        int n = 5, k = 2;
        Console.WriteLine("Value of C(" + n + "," + k
                          + ") is " + binomialCoeff(n, k));
    }
}

// This code is contributed by anuj_67.

Javascript

<script>

// A Dynamic Programming based
// solution that uses table C to
// calculate the Binomial Coefficient

// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff(n, k)
{
    var C = Array(n + 1).fill(0).map(
      x => Array(k + 1).fill(0));;
    var i, j;

    // Calculate  value of Binomial
    // Coefficient in bottom up manner
    for(i = 0; i <= n; i++)
    {
        for(j = 0; j <= min(i, k); j++)
        {

            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;

            // Calculate value using
            // previously stored values
            else
                C[i][j] = C[i - 1][j - 1] +
                          C[i - 1][j];
        }
    }
    return C[n][k];
}

// A utility function to return
// minimum of two integers
function min(a, b)
{
    return (a < b) ? a : b;
}

// Driver code
var n = 5, k = 2;
document.write("Value of C(" + n + "," + k +
                     ") is " + binomialCoeff(n, k));

// This code is contributed by 29AjayKumar

</script>

PHP

<?php
// A Dynamic Programming based
// solution that uses table C[][] to
// calculate the Binomial Coefficient

// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff( $n, $k)
{
    $C = array(array());
    $i; $j;

    // Calculate value of Binomial
    // Coefficient in bottom up manner
    for ($i = 0; $i <= $n; $i++)
    {
        for ($j = 0; $j <= min($i, $k); $j++)
        {

            // Base Cases
            if ($j == 0 || $j == $i)
                $C[$i][$j] = 1;

            // Calculate value using
            // previously stored values
            else
                $C[$i][$j] = $C[$i - 1][$j - 1] +
                                 $C[$i - 1][$j];
        }
    }

    return $C[$n][$k];
}

    // Driver Code
    $n = 5;
    $k = 2;
    echo "Value of C(" ,$n," ",$k, ") is"," "
                 , binomialCoeff($n, $k) ;

// This code is contributed by anuj_67.
?>

Output

Value of C[5][2] is 10

Time Complexity: O(n*k) 
Auxiliary Space: O(n*k)

Following is a space-optimized version of the above code. The following code only uses O(k). Thanks to AK for suggesting this method. 

C++

// C++ program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
#include <bits/stdc++.h>
using namespace std;

int binomialCoeff(int n, int k)
{
    int C[k + 1];
    memset(C, 0, sizeof(C));

    C[0] = 1; // nC0 is 1

    for (int i = 1; i <= n; i++)
    {

        // Compute next row of pascal triangle using
        // the previous row
        for (int j = min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}

/* Driver code*/
int main()
{
    int n = 5, k = 2;
    cout << "Value of C(" << n << "," << k << ")"<< "is " <<binomialCoeff(n, k);
    return 0;
}

// This code is contributed by shivanisinghss2110

C

// C program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
#include <stdio.h>

int binomialCoeff(int n, int k)
{
    int C[k + 1];
    memset(C, 0, sizeof(C));

    C[0] = 1; // nC0 is 1

    for (int i = 1; i <= n; i++) {
        // Compute next row of pascal triangle using
        // the previous row
        for (int j = min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}

/* Driver code*/
int main()
{
    int n = 5, k = 2;
    printf("Value of C(%d, %d) is %d ", n, k,
           binomialCoeff(n, k));
    return 0;
}

Java

// JAVA Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
import java.util.*;

class GFG {

    static int binomialCoeff(int n, int k)
    {
        int C[] = new int[k + 1];

        // nC0 is 1
        C[0] = 1;

        for (int i = 1; i <= n; i++) {
            // Compute next row of pascal
            // triangle using the previous row
            for (int j = Math.min(i, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }
        return C[k];
    }

    /* Driver code  */
    public static void main(String[] args)
    {
        int n = 5, k = 2;
        System.out.printf("Value of C(%d, %d) is %d ", n, k,
                          binomialCoeff(n, k));
    }
}

Python3

# Python program for Optimized
# Dynamic Programming solution to
# Binomial Coefficient. This one
# uses the concept of pascal
# Triangle and less memory


def binomialCoeff(n, k):

    # Declaring an empty array
    C = [0 for i in range(k+1)]
    C[0] = 1  # since nC0 is 1

    for i in range(1, n+1):

        # Compute next row of pascal triangle using
        # the previous row
        j = min(i, k)
        while (j > 0):
            C[j] = C[j] + C[j-1]
            j -= 1

    return C[k]


# Driver Code
n = 5
k = 2
print ("Value of C(%d,%d) is %d" % (n, k, binomialCoeff(n, k)))

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# Code for Dynamic Programming |
// Set 9 (Binomial Coefficient)
using System;

class GFG {

    static int binomialCoeff(int n, int k)
    {
        int[] C = new int[k + 1];

        // nC0 is 1
        C[0] = 1;

        for (int i = 1; i <= n; i++) {
            // Compute next row of pascal
            // triangle using the previous
            // row
            for (int j = Math.Min(i, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }
        return C[k];
    }

    /* Driver Code */
    public static void Main()
    {
        int n = 5, k = 2;
        Console.WriteLine("Value of C(" + n + " " + k
                          + ") is " + binomialCoeff(n, k));
    }
}

// This code is contributed by anuj_67.

Javascript

<script>

// Javascript program for space optimized
// Dynamic Programming
// Solution of Binomial Coefficient
function binomialCoeff(n, k)
{
    let C = new Array(k + 1);
    C.fill(0);

    // nC0 is 1
    C[0] = 1;

    for(let i = 1; i <= n; i++)
    {

        // Compute next row of pascal
        // triangle using the previous
        // row
        for(let j = Math.min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}

// Driver code
let n = 5, k = 2;
document.write("Value of C(" + n + " " +
               k + ") is " + binomialCoeff(n, k));

// This code is contributed by divyesh072019

</script>

PHP

<?php
// PHP program for space optimized
// Dynamic Programming Solution of
// Binomial Coefficient
function binomialCoeff($n, $k)
{
    $C = array_fill(0, $k + 1, 0);

    $C[0] = 1; // nC0 is 1

    for ($i = 1; $i <= $n; $i++)
    {
        // Compute next row of pascal
        // triangle using the previous row
        for ($j = min($i, $k); $j > 0; $j--)
            $C[$j] = $C[$j] + $C[$j - 1];
    }
    return $C[$k];
}

// Driver Code
$n = 5; $k = 2;
echo "Value of C[$n, $k] is ".
        binomialCoeff($n, $k);

// This code is contributed by mits.
?>

Output

Value of C(5, 2) is 10 

Time Complexity: O(n*k) 
Auxiliary Space: O(k)

Explanation:
1==========>> n = 0, C(0,0) = 1 
1–1========>> n = 1, C(1,0) = 1, C(1,1) = 1 
1–2–1======>> n = 2, C(2,0) = 1, C(2,1) = 2, C(2,2) = 1 
1–3–3–1====>> n = 3, C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C(3,3)=1 
1–4–6–4–1==>> n = 4, C(4,0) = 1, C(4,1) = 4, C(4,2) = 6, C(4,3)=4, C(4,4)=1 
So here every loop on i, builds i’th row of pascal triangle, using (i-1)th row
At any time, every element of array C will have some value (ZERO or more) and in the next iteration, the value for those elements comes from the previous iteration. 
In statement, 
C[j] = C[j] + C[j-1] 
The right-hand side represents the value coming from the previous iteration (A row of Pascal’s triangle depends on the previous row). The left-Hand side represents the value of the current iteration which will be obtained by this statement. 

Let's say we want to calculate C(4, 3), 
i.e. n=4, k=3:
All elements of array C of size 4 (k+1) are
initialized to ZERO.
i.e. C[0] = C[1] = C[2] = C[3] = C[4] = 0;
Then C[0] is set to 1
For i = 1:
C[1] = C[1] + C[0] = 0 + 1 = 1 ==>> C(1,1) = 1
For i = 2:
C[2] = C[2] + C[1] = 0 + 1 = 1 ==>> C(2,2) = 1
C[1] = C[1] + C[0] = 1 + 1 = 2 ==>> C(2,1) = 2
For i=3:
C[3] = C[3] + C[2] = 0 + 1 = 1 ==>> C(3,3) = 1
C[2] = C[2] + C[1] = 1 + 2 = 3 ==>> C(3,2) = 3
C[1] = C[1] + C[0] = 2 + 1 = 3 ==>> C(3,1) = 3
For i=4:
C[4] = C[4] + C[3] = 0 + 1 = 1 ==>> C(4,4) = 1
C[3] = C[3] + C[2] = 1 + 3 = 4 ==>> C(4,3) = 4
C[2] = C[2] + C[1] = 3 + 3 = 6 ==>> C(4,2) = 6
C[1] = C[1] + C[0] = 3 + 1 = 4 ==>> C(4,1) = 4
C(4,3) = 4 is would be the answer in our example.

Memoization Approach: The idea is to create a lookup table and follow the recursive top-down approach. Before computing any value, we check if it is already in the lookup table. If yes, we return the value. Else we compute the value and store it in the lookup table. Following is the Top-down approach of dynamic programming to finding the value of the Binomial Coefficient. 

C++

// A Dynamic Programming based
// solution that uses
// table dp[][] to calculate
// the Binomial Coefficient
// A naive recursive approach
// with table C++ implementation
#include <bits/stdc++.h>
using namespace std;

// Returns value of Binomial Coefficient C(n, k)
int binomialCoeffUtil(int n, int k, int** dp)
{
    // If value in lookup table then return
    if (dp[n][k] != -1) //
        return dp[n][k];

    // store value in a table before return
    if (k == 0) {
        dp[n][k] = 1;
        return dp[n][k];
    }

    // store value in table before return
    if (k == n) {
        dp[n][k] = 1;
        return dp[n][k];
    }

    // save value in lookup table before return
    dp[n][k] = binomialCoeffUtil(n - 1, k - 1, dp) +
               binomialCoeffUtil(n - 1, k, dp);
    return dp[n][k];
}

int binomialCoeff(int n, int k)
{
    int** dp; // make a temporary lookup table
    dp = new int*[n + 1];

    // loop to create table dynamically
    for (int i = 0; i < (n + 1); i++) {
        dp[i] = new int[k + 1];
    }

    // nested loop to initialise the table with -1
    for (int i = 0; i < (n + 1); i++) {
        for (int j = 0; j < (k + 1); j++) {
            dp[i][j] = -1;
        }
    }

    return binomialCoeffUtil(n, k, dp);
}

/* Driver code*/
int main()
{
    int n = 5, k = 2;
    cout << "Value of C(" << n << ", " << k << ") is "
         << binomialCoeff(n, k) << endl;
    return 0;
}

// This is code is contributed by MOHAMMAD MUDASSIR

Java

// A Dynamic Programming based
// solution that uses
// table dp[][] to calculate
// the Binomial Coefficient
// A naive recursive approach
// with table Java implementation
import java.util.*;
class GFG{

// Returns value of Binomial
// Coefficient C(n, k)
static int binomialCoeffUtil(int n, int k,
                             Vector<Integer> []dp)
{
  // If value in lookup table
  // then return
  if (dp[n].get(k) != -1)
    return dp[n].get(k);

  // store value in a table
  // before return
  if (k == 0)
  {
    dp[n].add(k, 1);
    return dp[n].get(k);
  }

  // store value in table
  // before return
  if (k == n)
  {
    dp[n].add(k, 1);
    return dp[n].get(k);
  }

  // save value in lookup table
  // before return
  dp[n].add(k, binomialCoeffUtil(n - 1,
                                 k - 1, dp) +
               binomialCoeffUtil(n - 1,
                                 k, dp));
  return dp[n].get(k);
}

static int binomialCoeff(int n, int k)
{
  // Make a temporary lookup table
  Vector<Integer> []dp = new Vector[n+1];

  // Loop to create table dynamically
  for (int i = 0; i < (n + 1); i++)
  {
    dp[i] = new Vector<Integer>();
    for(int j = 0; j <= k; j++)
      dp[i].add(-1);
  }
  return binomialCoeffUtil(n, k, dp);
}

// Driver code
public static void main(String[] args)
{
  int n = 5, k = 2;
  System.out.print("Value of C(" + n +
                   ", " + k + ") is " +
                   binomialCoeff(n, k) + "\n");
}
}

// This code is contributed by Rajput-Ji

Python3

# A Dynamic Programming based solution
# that uses table dp[][] to calculate
# the Binomial Coefficient. A naive
# recursive approach with table
# Python3 implementation

# Returns value of Binomial
# Coefficient C(n, k)
def binomialCoeffUtil(n, k, dp):

    # If value in lookup table then return
    if dp[n][k] != -1:
        return dp[n][k]

    # Store value in a table before return
    if k == 0:
        dp[n][k] = 1
        return dp[n][k]

    # Store value in table before return
    if k == n:
        dp[n][k] = 1
        return dp[n][k]

    # Save value in lookup table before return
    dp[n][k] = (binomialCoeffUtil(n - 1, k - 1, dp) +
                binomialCoeffUtil(n - 1, k, dp))

    return dp[n][k]

def binomialCoeff(n, k):

    # Make a temporary lookup table
    dp = [ [ -1 for y in range(k + 1) ]
                for x in range(n + 1) ]

    return binomialCoeffUtil(n, k, dp)

# Driver code
n = 5
k = 2

print("Value of C(" + str(n) +
               ", " + str(k) + ") is",
               binomialCoeff(n, k))

# This is code is contributed by Prateek Gupta

C#

// C# program for the
// above approach

// A Dynamic Programming based
// solution that uses
// table [,]dp to calculate
// the Binomial Coefficient
// A naive recursive approach
// with table C# implementation
using System;
using System.Collections.Generic;
class GFG{

// Returns value of Binomial
// Coefficient C(n, k)
static int binomialCoeffUtil(int n, int k,
                             List<int> []dp)
{
  // If value in lookup table
  // then return
  if (dp[n][k] != -1)
    return dp[n][k];

  // store value in a table
  // before return
  if (k == 0)
  {
    dp[n][k] = 1;
    return dp[n][k];
  }

  // store value in table
  // before return
  if (k == n)
  {
    dp[n][k] = 1;
    return dp[n][k];
  }

  // save value in lookup table
  // before return
  dp[n][k] = binomialCoeffUtil(n - 1,
                               k - 1, dp) +
             binomialCoeffUtil(n - 1,
                               k, dp);
  return dp[n][k];
}

static int binomialCoeff(int n, int k)
{
  // Make a temporary lookup table
  List<int> []dp = new List<int>[n + 1];

  // Loop to create table dynamically
  for (int i = 0; i < (n + 1); i++)
  {
    dp[i] = new List<int>();

    for(int j = 0; j <= k; j++)
      dp[i].Add(-1);
  }
  return binomialCoeffUtil(n, k, dp);
}

// Driver code
public static void Main(String[] args)
{
  int n = 5, k = 2;
  Console.Write("Value of C(" + n +
                ", " + k + ") is " +
                binomialCoeff(n, k) + "\n");
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// A Dynamic Programming based solution that
// uses table dp[][] to calculate the
// Binomial Coefficient. A naive recursive
// approach with table Javascript implementation

// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeffUtil(n, k, dp)
{

    // If value in lookup table
    // then return
    if (dp[n][k] != -1)
        return dp[n][k];

    // Store value in a table
    // before return
    if (k == 0)
    {
        dp[n][k] = 1;
        return dp[n][k];
    }

    // Store value in table
    // before return
    if (k == n)
    {
        dp[n][k] = 1;
        return dp[n][k];
    }

    // Save value in lookup table
    // before return
    dp[n][k] = binomialCoeffUtil(n - 1,
                                 k - 1, dp) +
               binomialCoeffUtil(n - 1,
                                 k, dp);
    return dp[n][k];
}

function binomialCoeff(n, k)
{

    // Make a temporary lookup table
    let dp = new Array(n + 1);

    // Loop to create table dynamically
    for(let i = 0; i < (n + 1); i++)
    {
        dp[i] = [];
        for(let j = 0; j <= k; j++)
            dp[i].push(-1);
    }
    return binomialCoeffUtil(n, k, dp);
}

// Driver code
let n = 5, k = 2;
document.write("Value of C(" + n +
               ", " + k + ") is " +
               binomialCoeff(n, k) + "\n");

// This code is contributed by avanitrachhadiya2155

</script>

 Time Complexity: O(n*k)

Auxiliary Space: O(n*k)

Output

Value of C(5, 2) is 10

Cancellation of factors between numerator and denominator:

nCr = (n-r+1)*(n-r+2)*….*n / (r!)

Create an array arr of numbers from n-r+1 to n which will be of size r. As nCr is always an integer, all numbers in the denominator should cancel with the product of the numerator (represented by arr).

for i = 1 to i = r,

        search arr, if arr[j] and i have gcd>1, divide both by the gcd and when i becomes 1, stop the search

Now, the answer is just the product of arr, whose value mod 10^9+7 can be found using a single pass and the formula use (a*b)%mod = (a%mod * b%mod)%mod 

C++

// C++ program to find gcd of
// two numbers in O(log(min(a,b)))

#include <bits/stdc++.h>
using namespace std;

int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

int nCr(int n, int r)
{
    // base case
    if (r > n)
        return 0;

    // C(n,r) = C(n,n-r)
    if (r > n - r)
        r = n - r;

    int mod = 1000000007;

    // array of elements from n-r+1 to n
    int arr[r];

    for (int i = n - r + 1; i <= n; i++) {
        arr[i + r - n - 1] = i;
    }

    long int ans = 1;
    // for numbers from 1 to r find arr[j],
    // such that gcd(i,arr[j])>1
    for (int k = 1; k < r + 1; k++) {
        int j = 0, i = k;
        while (j < r) {
            int x = gcd(i, arr[j]);
            if (x > 1) {
                // if gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }

            // if i becomes 1, no need to search arr
            if (i == 1)
                break;
            j += 1;
        }
    }

    // single pass to multiply the numerator
    for (int i : arr)
        ans = (ans * i) % mod;
    return (int)ans;
}

int main()
{
    int n = 5, r = 2;
    cout << "Value of C(" << n << ", " << r << ") is "
         << nCr(n, r) << "\n";
    return 0;
}

// This code is contributed by rajsanghavi9.

Java

import java.util.*;
class GFG
{
    static int gcd(int a, int b) // function to find gcd of two numbers in O(log(min(a,b)))
    {
        if(b==0) return a;
        return gcd(b,a%b);
    }
    static int nCr(int n, int r)
    {
        if(r>n) // base case
            return 0;
        if(r>n-r) // C(n,r) = C(n,n-r) better time complexity for lesser r value
            r = n-r;
        int mod = 1000000007;
        int[] arr = new int[r]; // array of elements from n-r+1 to n
        for(int i=n-r+1; i<=n; i++)
        {
            arr[i+r-n-1] = i;
        }
        long ans = 1;
        for(int k=1;k<r+1;k++) // for numbers from 1 to r find arr[j] such that gcd(i,arr[j])>1
        {
            int j=0, i=k;
            while(j<arr.length)
            {
                int x = gcd(i,arr[j]);
                if(x>1)
                {
                    arr[j] /= x; // if gcd>1, divide both by gcd
                    i /= x;
                }
                if(i==1)
                    break; // if i becomes 1, no need to search arr
                j += 1;
            }
        }
        for(int i : arr) // single pass to multiply the numerator
            ans = (ans*i)%mod;
        return (int)ans;
    }
    // Driver code
    public static void main(String[] args)
    {
        int n = 5, r = 2;
        System.out.print("Value of C(" +  n+ ", " +  r+ ") is "
             +nCr(n, r) +"\n");
    }
}
// This code is contributed by Gautam Wadhwani

Python3

import math
class GFG:
    def nCr(self, n, r):
        def gcd(a,b): # function to find gcd of two numbers in O(log(min(a,b)))
            if b==0: # base case
                return a
            return gcd(b,a%b)
        if r>n:
            return 0
        if r>n-r: # C(n,r) = C(n,n-r) better time complexity for lesser r value
            r = n-r
        mod = 10**9 + 7
        arr = list(range(n-r+1,n+1)) # array of elements from n-r+1 to n
        ans = 1
        for i in range(1,r+1): # for numbers from 1 to r find arr[j] such that gcd(i,arr[j])>1
            j=0
            while j<len(arr):
                x = gcd(i,arr[j])
                if x>1:
                    arr[j] //= x # if gcd>1, divide both by gcd
                    i //= x
                if arr[j]==1: # if element becomes 1, its of no use anymore so delete from arr
                    del arr[j]
                    j -= 1
                if i==1:
                    break # if i becomes 1, no need to search arr
                j += 1
        for i in arr: # single pass to multiply the numerator
            ans = (ans*i)%mod
        return ans
     # Driver code
n = 5
k = 2
ob = GFG()
print("Value of C(" + str(n) +
               ", " + str(k) + ") is",
               ob.nCr(n, k))

# This is code is contributed by Gautam Wadhwani

C#

using System;

class GFG{

// Function to find gcd of two numbers
// in O(log(min(a,b)))
static int gcd(int a, int b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

static int nCr(int n, int r)
{

    // Base case
    if (r > n)
        return 0;

    // C(n,r) = C(n,n-r) better time
    // complexity for lesser r value
    if (r > n - r)
        r = n - r;

    int mod = 1000000007;

    // Array of elements from n-r+1 to n
    int[] arr = new int[r];
    for(int i = n - r + 1; i <= n; i++)
    {
        arr[i + r - n - 1] = i;
    }
    long ans = 1;

    // For numbers from 1 to r find arr[j]
    // such that gcd(i,arr[j])>1
    for(int k = 1; k < r + 1; k++)
    {
        int j = 0, i = k;
        while (j < arr.Length)
        {
            int x = gcd(i,arr[j]);
            if (x > 1)
            {

                // If gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }

            if (i == 1)

                // If i becomes 1, no need
                // to search arr
                break;

            j += 1;
        }
    }

    // Single pass to multiply the numerator
    foreach(int i in arr)
        ans = (ans * i) % mod;

    return (int)ans;
}

// Driver code
static public void Main()
{
    int n = 5, r = 2;
    Console.WriteLine("Value of C(" +  n +
                      ", " +  r + ") is " +
                      nCr(n, r) + "\n");
}
}

// This code is contributed by rag2127

Javascript

<script>

// Javascript program to find gcd of
// two numbers in O(log(min(a,b)))
function gcd(a, b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

function nCr(n, r)
{

    // Base case
    if (r > n)
        return 0;

    // C(n,r) = C(n,n-r) better time
    // complexity for lesser r value
    if (r > n - r)
        r = n - r;

     mod = 1000000007;

    // Array of elements from n-r+1 to n
    var arr = new Array(r);
    for(var i = n - r + 1; i <= n; i++)
    {
        arr[i + r - n - 1] = i;
    }
    var ans = 1;

    // For numbers from 1 to r find arr[j]
    // such that gcd(i,arr[j])>1
    for(var k = 1; k < r + 1 ; k++)
    {
        var j = 0, i = k;
        while (j < arr.length)
        {
            var x = gcd(i, arr[j]);
            if (x > 1)
            {

                // If gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }

            // If i becomes 1, no need to search arr
            if (i == 1)
                break;

            j += 1;
        }
    }

    // Single pass to multiply the numerator
    arr.forEach(function (i)
    {
        ans = (ans * i) % mod;
    });

    return ans;
}

// Driver code
var n = 5, r = 2;
document.write("Value of C(" + n + ", " +
               r + ") is " + nCr(n, r) + "\n");

// This code is contributed by shivanisinghss2110

</script>

Output

Value of C(5, 2) is 10

Time Complexity: O(( min(r, n-r)^2 ) * log(n)),   useful when n >> r  or  n >> (n-r)

Auxiliary Space: O(min(r, n-r))

See this for GCD in logarithmic time

Prime factorization of every number from 1 to n using Sieve of Eratosthenes :

1. Create an array SPF of size n+1 to the smallest prime factor of each number from 1 to n 

Set SPF[i] = i for all i = 1 to i = n

2. Use Sieve of Eratosthenes:

for i = 2 to i = n:
    if i is prime,
       for all multiples j of i, j<=n:
           if SPF[j] equals j, set SPF[j] = i

3. Once, we know the SPF of each number from 1 to n, we can find the prime factorization of any number from 1 to n in O(log(n)) time using recursive division by SPF until the number becomes 1

Now, nCr  =  (n-r+1)*(r+2)* ... *(n) / (r)!

4. Create a dictionary (or hashmap) to store the frequency of each prime in the prime factorization of the actual value of nCr.

5. So, just calculate the frequency of each prime in nCr and multiply them raised to the power of their frequency.

6. For the numerator, iterate through i = n-r+1 to i = n, and for all prime factors of i, store their frequency in a dictionary.

( prime_pow[prime_factor] += freq_in_i ) 

7. For the denominator, iterate through i = 1 to i = r and now subtract the frequency instead of adding.

8. Now, traverse the dictionary and multiply the answer to (prime ^ prime_pow[prime]) % (10^9 + 7)

ans = (ans * pow(prime, prime_pow[prime], mod) ) % mod 

C++

#include <bits/stdc++.h>
using namespace std;

// pow(base,exp,mod) is used to find
// (base^exp)%mod fast -> O(log(exp))
long int pow(long int b, long int exp, long int mod)
{
    long int ret = 1;

    while (exp > 0) {
        if ((exp & 1) > 0)
            ret = (ret * b) % mod;
        b = (b * b) % mod;
        exp >>= 1;
    }

    return ret;
}

int nCr(int n, int r)
{
    // base case
    if (r > n)
        return 0;

    // C(n,r) = C(n,n-r) Complexity for
    // this code is lesser for lower n-r
    if (n - r > r)
        r = n - r;

    // list to smallest prime factor
    // of each number from 1 to n
    int SPF[n + 1];

    // set smallest prime factor of each
    // number as itself
    for (int i = 1; i <= n; i++)
        SPF[i] = i;

    // set smallest prime factor of all
    // even numbers as 2
    for (int i = 4; i <= n; i += 2)
        SPF[i] = 2;

    for (int i = 3; i * i < n + 1; i += 2) {

        // Check if i is prime
        if (SPF[i] == i) {
            // All multiples of i are
            // composite (and divisible by
            // i) so add i to their prime
            // factorization getpow(j,i)
            // times
            for (int j = i * i; j < n + 1; j += i)
                if (SPF[j] == j) {
                    SPF[j] = i;
                }
        }
    }
    // Hash Map to store power of
    // each prime in C(n,r)
    map<int, int> prime_pow;

    // For numerator count frequency of each prime factor
    for (int i = r + 1; i < n + 1; i++) {

        int t = i;

        // Recursive division to find
        // prime factorization of i
        while (t > 1) {
            if (!prime_pow[SPF[t]]) {
                prime_pow[SPF[t]] = 1;
            }
            else
                prime_pow[SPF[t]]++;
            // prime_pow.put(SPF[t],
            // prime_pow.getOrDefault(SPF[t], 0)
            // + 1);
            t /= SPF[t];
        }
    }

    // For denominator subtract the power of
    // each prime factor
    for (int i = 1; i < n - r + 1; i++) {
        int t = i;

        // Recursive division to find
        // prime factorization of i
        while (t > 1) {
            prime_pow[SPF[t]]--;
            // prime_pow.put(SPF[t],
            // prime_pow.get(SPF[t]) - 1);
            t /= SPF[t];
        }
    }

    // long because mod is large and a%mod
    // * b%mod can overflow int
    long int ans = 1, mod = 1000000007;

    // use (a*b)%mod = (a%mod * b%mod)%mod
    for (auto it : prime_pow)

        // pow(base,exp,mod) is used to
        // find (base^exp)%mod fast
        ans = (ans
               * pow(it.first, prime_pow[it.first], mod))
              % mod;
    return (int)ans;
}

int main()
{
    int n = 5, r = 2;
    cout << "Value of C(" << n << ", " << r << ") is "
         << nCr(n, r) << "\n";
    return 0;
}

// This code is contributed by rajsanghavi9.

Java

import java.util.*;
class GFG {

    // pow(base,exp,mod) is used to find
    // (base^exp)%mod fast -> O(log(exp))
    static long pow(long b, long exp, long mod)
    {
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) > 0)
                ret = (ret * b) % mod;
            b = (b * b) % mod;
            exp >>= 1;
        }
        return ret;
    }
    static int nCr(int n, int r)
    {
        if (r > n) // base case
            return 0;

        // C(n,r) = C(n,n-r) Complexity for
        // this code is lesser for lower n-r
        if (n - r > r)
            r = n - r;

        // list to smallest prime factor
        // of each number from 1 to n
        int[] SPF = new int[n + 1];

        // set smallest prime factor of each
        // number as itself
        for (int i = 1; i <= n; i++)
            SPF[i] = i;

        // set smallest prime factor of all
        // even numbers as 2
        for (int i = 4; i <= n; i += 2)
            SPF[i] = 2;

        for (int i = 3; i * i < n + 1; i += 2)
        {

            // Check if i is prime
            if (SPF[i] == i)
            {

                // All multiples of i are
                // composite (and divisible by
                // i) so add i to their prime
                // factorization getpow(j,i)
                // times
                for (int j = i * i; j < n + 1; j += i)
                    if (SPF[j] == j) {
                        SPF[j] = i;
                    }
            }
        }

       // Hash Map to store power of
       // each prime in C(n,r)
       Map<Integer, Integer> prime_pow
            = new HashMap<>();

        // For numerator count frequency of each prime factor
        for (int i = r + 1; i < n + 1; i++)
        {
            int t = i;

            // Recursive division to find
            // prime factorization of i
            while (t > 1)
            {
                prime_pow.put(SPF[t],
                    prime_pow.getOrDefault(SPF[t], 0) + 1);
                t /= SPF[t];
            }
        }

        // For denominator subtract the power of
        // each prime factor
        for (int i = 1; i < n - r + 1; i++)
        {
            int t = i;

            // Recursive division to find
            // prime factorization of i
            while (t > 1)
            {
                prime_pow.put(SPF[t],
                              prime_pow.get(SPF[t]) - 1);
                t /= SPF[t];
            }
        }

        // long because mod is large and a%mod
        // * b%mod can overflow int
        long ans = 1, mod = 1000000007;

        // use (a*b)%mod = (a%mod * b%mod)%mod
        for (int i : prime_pow.keySet())

            // pow(base,exp,mod) is used to
            // find (base^exp)%mod fast
            ans = (ans * pow(i, prime_pow.get(i), mod))
                  % mod;
        return (int)ans;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 5, r = 2;
        System.out.print("Value of C(" + n + ", " + r
                         + ") is " + nCr(n, r) + "\n");
    }
}
// This code is contributed by Gautam Wadhwani

Python3

# Python code for the above approach
import math

class GFG:
    def nCr(self, n, r):

        # Base case
        if r > n:
            return 0

        # C(n,r) = C(n,n-r) Complexity for this
        # code is lesser for lower n-r
        if n - r > r:
            r = n - r

        # List to store smallest prime factor
        # of every number from 1 to n
        SPF = [i for i in range(n+1)]
        for i in range(4, n+1, 2):

            # set smallest prime factor of
            # all even numbers as 2
            SPF[i] = 2

        for i in range(3, n+1, 2):

            if i*i > n:
                break

            # Check if i is prime
            if SPF[i] == i:

                # All multiples of i are composite
                # (and divisible by i) so add i to
                # their prime factorization getpow(j,i) times
                for j in range(i*i, n+1, i):
                    if SPF[j] == j:

                        # set smallest prime factor
                        # of j to i only if it is
                        # not previously set
                        SPF[j] = i

         # dictionary to store power of each prime in C(n,r)
        prime_pow = {}

        # For numerator count frequency
        # of each prime factor
        for i in range(r+1, n+1):
            t = i

            # Recursive division to
            # find prime factorization of i
            while t > 1:
                if not SPF[t] in prime_pow:
                    prime_pow[SPF[t]] = 1
                else:
                    prime_pow[SPF[t]] += 1
                t //= SPF[t]

        # For denominator subtract the
        # power of each prime factor
        for i in range(1, n-r+1):
            t = i

            # Recursive division to
            # find prime factorization of i
            while t > 1:
                prime_pow[SPF[t]] -= 1
                t //= SPF[t]
        ans = 1
        mod = 10**9 + 7

         # Use (a*b)%mod = (a%mod * b%mod)%mod
        for i in prime_pow:

            # pow(base,exp,mod) is used to
            # find (base^exp)%mod fast
            ans = (ans*pow(i, prime_pow[i], mod)) % mod
        return ans


# Driver code
n = 5
k = 2
ob = GFG()
print("Value of C(" + str(n) +
      ", " + str(k) + ") is",
      ob.nCr(n, k))

# This is code is contributed by Gautam Wadhwani

C#

using System;
using System.Collections.Generic;

public class GFG {

    // pow(base,exp,mod) is used to find
    // (base^exp)%mod fast -> O(log(exp))
    static long pow(long b, long exp, long mod)
    {
        long ret = 1;
        while (exp > 0) {
            if ((exp & 1) > 0)
                ret = (ret * b) % mod;
            b = (b * b) % mod;
            exp >>= 1;
        }
        return ret;
    }
    static int nCr(int n, int r)
    {
        if (r > n) // base case
            return 0;

        // C(n,r) = C(n,n-r) Complexity for
        // this code is lesser for lower n-r
        if (n - r > r)
            r = n - r;

        // list to smallest prime factor
        // of each number from 1 to n
        int[] SPF = new int[n + 1];

        // set smallest prime factor of each
        // number as itself
        for (int i = 1; i <= n; i++)
            SPF[i] = i;

        // set smallest prime factor of all
        // even numbers as 2
        for (int i = 4; i <= n; i += 2)
            SPF[i] = 2;

        for (int i = 3; i * i < n + 1; i += 2) {

            // Check if i is prime
            if (SPF[i] == i) {

                // All multiples of i are
                // composite (and divisible by
                // i) so add i to their prime
                // factorization getpow(j,i)
                // times
                for (int j = i * i; j < n + 1; j += i)
                    if (SPF[j] == j) {
                        SPF[j] = i;
                    }
            }
        }

        // Hash Map to store power of
        // each prime in C(n,r)
        Dictionary<int, int> prime_pow
            = new Dictionary<int, int>();

        // For numerator count frequency of each prime
        // factor
        for (int i = r + 1; i < n + 1; i++) {
            int t = i;

            // Recursive division to find
            // prime factorization of i
            while (t > 1) {
                if (prime_pow.ContainsKey(SPF[t])) {
                    prime_pow[SPF[t]]
                        = prime_pow[SPF[t]] + 1;
                }
                else {
                    prime_pow.Add(SPF[t], 1);
                }
                t /= SPF[t];
            }
        }

        // For denominator subtract the power of
        // each prime factor
        for (int i = 1; i < n - r + 1; i++) {
            int t = i;

            // Recursive division to find
            // prime factorization of i
            while (t > 1) {

                if (prime_pow.ContainsKey(SPF[t])) {
                    prime_pow[SPF[t]]
                        = prime_pow[SPF[t]] - 1;
                }

                t /= SPF[t];
            }
        }

        // long because mod is large and a%mod
        // * b%mod can overflow int
        long ans = 1, mod = 1000000007;

        // use (a*b)%mod = (a%mod * b%mod)%mod
        foreach(int i in prime_pow.Keys)

            // pow(base,exp,mod) is used to
            // find (base^exp)%mod fast
            ans
            = (ans * pow(i, prime_pow[i], mod)) % mod;
        return (int)ans;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int n = 5, r = 2;
        Console.Write("Value of C(" + n + ", " + r + ") is "
                      + nCr(n, r) + "\n");
    }
}

// This code contributed by gauravrajput1

Javascript

<script>

// Javascript program to find gcd of
// two numbers in O(log(min(a,b)))

function gcd(a, b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

function nCr(n, r)
{
    // base case
    if (r > n)
        return 0;

    // C(n,r) = C(n,n-r)
    if (r > n - r)
        r = n - r;

    var mod = 1000000007;

    // array of elements from n-r+1 to n
    var arr = new Array(r);

    for (var i = n - r + 1; i <= n; i++) {
        arr[i + r - n - 1] = i;
    }

    var ans = 1;
    // for numbers from 1 to r find arr[j],
    // such that gcd(i,arr[j])>1
    for (var k = 1; k < r + 1; k++) {
        var j = 0;
        var i = k;
        do {
            var x = gcd(i, arr[j]);
            if (x > 1) {
                // if gcd>1, divide both by gcd
                arr[j] /= x;
                i /= x;
            }

            // if i becomes 1, no need to search arr
            if (i == 1)
                break;
            j += 1;
        }
        while (j < r);
    }

    // single pass to multiply the numerator

    arr.forEach(function (i, index) {
          ans = (ans * i) % mod;
    });
    return ans;
}


var n = 5;
var r = 2;
document.write("Value of C(" + n + ", " + r + ") is " + nCr(n, r) + "<br>");

// This code is contributed by shivani.

</script>

Output

Value of C(5, 2) is 10

Time Complexity: O(n*log(n)),  so useful when r->n/2

Auxiliary Space: O(n)

See this for Prime factorization in O(log(n))

Another Approach: (Modular Inversion technique)

1. The general formula of nCr is ( n*(n-1)*(n-2)* … *(n-r+1) ) / (r!). We can directly use this formula to find nCr. But that will overflow out of bound. We need to find nCr mod m so that it doesn’t overflow. We can easily do it with modular arithmetic formula. 

for the  n*(n-1)*(n-2)* ... *(n-r+1) part we can use the formula,
(a*b) mod m = ((a % m) * (b % m)) % m

2. and for the 1/r! part, we need to find the modular inverse of every number from 1 to r. Then use the same formula above with a modular inverse of 1 to r. We can find modular inverse in O(r) time using  the formula, 

inv[1] = 1
inv[i] = − ⌊m/i⌋ * inv[m mod i] mod m
To use this formula, m has to be a prime.

In the practice problem, we need to show the answer with modulo 1000000007 which is a prime. 

So, this technique will work.

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find binomial
// coefficient
int binomialCoeff(int n, int r)
{

    if (r > n)
        return 0;
    long long int m = 1000000007;
    long long int inv[r + 1] = { 0 };
    inv[0] = 1;
    if(r+1>=2)
    inv[1] = 1;

    // Getting the modular inversion
    // for all the numbers
    // from 2 to r with respect to m
    // here m = 1000000007
    for (int i = 2; i <= r; i++) {
        inv[i] = m - (m / i) * inv[m % i] % m;
    }

    int ans = 1;

    // for 1/(r!) part
    for (int i = 2; i <= r; i++) {
        ans = ((ans % m) * (inv[i] % m)) % m;
    }

    // for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for (int i = n; i >= (n - r + 1); i--) {
        ans = ((ans % m) * (i % m)) % m;
    }
    return ans;
}

/* Driver code*/
int main()
{
    int n = 5, r = 2;
    cout << "Value of C(" << n << ", " << r << ") is "
         << binomialCoeff(n, r) << endl;
    return 0;
}

Java

// JAVA program for the above approach
import java.util.*;
class GFG
{

// Function to find binomial
// coefficient
static int binomialCoeff(int n, int r)
{

    if (r > n)
    return 0;
    long  m = 1000000007;
    long  inv[] = new long[r + 1];
    inv[0] = 1;
    if(r+1>=2)
    inv[1] = 1;

    // Getting the modular inversion
    // for all the numbers
    // from 2 to r with respect to m
    // here m = 1000000007
    for (int i = 2; i <= r; i++) {
        inv[i] = m - (m / i) * inv[(int) (m % i)] % m;
    }

    int ans = 1;

    // for 1/(r!) part
    for (int i = 2; i <= r; i++) {
        ans = (int) (((ans % m) * (inv[i] % m)) % m);
    }

    // for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for (int i = n; i >= (n - r + 1); i--) {
        ans = (int) (((ans % m) * (i % m)) % m);
    }
    return ans;
}

/* Driver code*/
public static void main(String[] args)
{
    int n = 5, r = 2;
    System.out.print("Value of C(" +  n+ ", " +  r+ ") is "
         +binomialCoeff(n, r) +"\n");
}
}

// This code contributed by Rajput-Ji

Python3

# Python3 program for the above approach

# Function to find binomial
# coefficient
def binomialCoeff(n, r):

    if (r > n):
        return 0

    m = 1000000007
    inv = [0 for i in range(r + 1)]
    inv[0] = 1
    if(r+1>=2):
        inv[1] = 1

    # Getting the modular inversion
    # for all the numbers
    # from 2 to r with respect to m
    # here m = 1000000007
    for i in range(2, r + 1):
        inv[i] = m - (m // i) * inv[m % i] % m

    ans = 1

    # for 1/(r!) part
    for i in range(2, r + 1):
        ans = ((ans % m) * (inv[i] % m)) % m

    # for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for i in range(n, n - r, -1):
        ans = ((ans % m) * (i % m)) % m

    return ans

# Driver code
n = 5
r = 2

print("Value of C(",n , "," , r ,") is ",binomialCoeff(n, r))

# This code is contributed by Pushpesh Raj

C#

// C# program for the above approach
using System;

public class GFG
{

// Function to find binomial
// coefficient
static int binomialCoeff(int n, int r)
{

    if (r > n)
    return 0;
    long  m = 1000000007;
    long  []inv = new long[r + 1];
    inv[0] = 1;
    if(r+1>=2)
    inv[1] = 1;

    // Getting the modular inversion
    // for all the numbers
    // from 2 to r with respect to m
    // here m = 1000000007
    for (int i = 2; i <= r; i++) {
        inv[i] = m - (m / i) * inv[(int) (m % i)] % m;
    }

    int ans = 1;

    // for 1/(r!) part
    for (int i = 2; i <= r; i++) {
        ans = (int) (((ans % m) * (inv[i] % m)) % m);
    }

    // for (n)*(n-1)*(n-2)*...*(n-r+1) part
    for (int i = n; i >= (n - r + 1); i--) {
        ans = (int) (((ans % m) * (i % m)) % m);
    }
    return ans;
}

/* Driver code*/
public static void Main(String[] args)
{
    int n = 5, r = 2;
    Console.Write("Value of C(" +  n+ ", " +  r+ ") is "
         +binomialCoeff(n, r) +"\n");
}
}



// This code is contributed by 29AjayKumar

Javascript

<script>
        // JavaScript Program for the above approach

        // Function to find binomial
        // coefficient
        function binomialCoeff(n, r) {

            if (r > n)
                return 0;
            let m = 1000000007;
            let inv = new Array(r + 1).fill(0);
            inv[0] = 1;
            if (r + 1 >= 2)
                inv[1] = 1;

            // Getting the modular inversion
            // for all the numbers
            // from 2 to r with respect to m
            // here m = 1000000007
            for (let i = 2; i <= r; i++) {
                inv[i] = m - Math.floor(m / i) * inv[m % i] % m;
            }

            let ans = 1;

            // for 1/(r!) part
            for (let i = 2; i <= r; i++) {
                ans = ((ans % m) * (inv[i] % m)) % m;
            }

            // for (n)*(n-1)*(n-2)*...*(n-r+1) part
            for (let i = n; i >= (n - r + 1); i--) {
                ans = ((ans % m) * (i % m)) % m;
            }
            return ans;
        }

        /* Driver code*/
        let n = 5, r = 2;
        document.write("Value of C(" + n + ", " + r + ") is "
            + binomialCoeff(n, r) + "<br>");

    // This code is contributed by Potta Lokesh
    </script>

Output

Value of C(5, 2) is 10

Time Complexity: O(n+k)

Auxiliary Space: O(k) 

See this for Space and time-efficient Binomial Coefficient

References: 
http://www.csl.mtu.edu/cs4321/www/Lectures/Lecture%2015%20-%20Dynamic%20Programming%20Binomial%20Coefficients.htm

https://cp-algorithms.com/algebra/module-inverse.html

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