Calculate nCr using Pascal’s Triangle

A useful application of Pascal’s triangle is the calculation of combinations. The formula to find ⁿC_r is n! / r! * (n – r)! which is also the formula for a cell of Pascal’s triangle. 
Pascal’s triangle:

Input: n = 5, r = 3
Output: 10
Explanation:
n! / r! * (n - r)! = 5! / 3! * (2!) = 120 / 12 = 10

Input: n = 7, r = 2
Output: 21
Explanation:
n! / r! * (n - r)! = 7! / 5! * (2!) = 42 / 2 = 21
 

Approach: The idea is to store the Pascal’s triangle in a matrix then the value of ⁿC_r will be the value of the cell at n^th row and r^th column. 
To create the pascal triangle use these two formula:

1. ⁿC₀ = 1, number of ways to select 0 elements from a set of n elements is 0
2. ⁿC_r = ^n-1C_r-1 + ^n-1C_r, number of ways to select r elements from a set of n elements is summation of ways to select r-1 elements from n-1 elements and ways to select r elements from n-1 elements.

The idea is to use the values of subproblems to calculate the answers for larger values. For example, to calculate ⁿC_r, use the values of ^n-1C_r-1 and ^n-1C_r. So DP can be used to preprocess all the values in the range.
Algorithm:

1. Create a matrix of size 1000 * 1000, assign the value of base cases, i.e. run a loop from 0 to 1000 and assign matrix[i][0] = 1, ⁿC₀ = 1
2. Run a nested loop from i = 1 to 1000 (outer loop) and the inner loop runs from j = 1 to i + 1.
3. For every element (i, j) assign the value of matrix[i][j] = matrix[i-1][j-1] + matrix[i-1][j], using the formula ⁿC_r = ^n-1C_r-1 + ^n-1C_r
4. After filling the matrix return the value of ⁿC_r as matrix[n][r]

Implementation:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Initialize the matrix with 0
int l[1001][1001] = { 0 };

void initialize()
{

    // 0C0 = 1
    l[0][0] = 1;
    for (int i = 1; i < 1001; i++) {
        // Set every nCr = 1 where r = 0
        l[i][0] = 1;
        for (int j = 1; j < i + 1; j++) {

            // Value for the current cell of Pascal's triangle
            l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]);
        }
    }
}

// Function to return the value of nCr
int nCr(int n, int r)
{
    // Return nCr
    return l[n][r];
}

// Driver code
int main()
{
    // Build the Pascal's triangle
    initialize();
    int n = 8;
    int r = 3;
    cout << nCr(n, r);
}

// This code is contributed by ihritik

Java

// Java implementation of the approach

class GFG {
    // Initialize the matrix with 0
    static int l[][] = new int[1001][1001];

    static void initialize()
    {

        // 0C0 = 1
        l[0][0] = 1;
        for (int i = 1; i < 1001; i++) {
            // Set every nCr = 1 where r = 0
            l[i][0] = 1;
            for (int j = 1; j < i + 1; j++) {

                // Value for the current cell of Pascal's triangle
                l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]);
            }
        }
    }

    // Function to return the value of nCr
    static int nCr(int n, int r)
    {
        // Return nCr
        return l[n][r];
    }
    // Driver code
    public static void main(String[] args)
    {
        // Build the Pascal's triangle
        initialize();
        int n = 8;
        int r = 3;
        System.out.println(nCr(n, r));
    }
}

// This code is contributed by ihritik

Python3

# Python3 implementation of the approach

# Initialize the matrix with 0
l = [[0 for i in range(1001)] for j in range(1001)]

def initialize():

    # 0C0 = 1
    l[0][0] = 1
    for i in range(1, 1001):

        # Set every nCr = 1 where r = 0
        l[i][0] = 1
        for j in range(1, i + 1):

            # Value for the current cell of Pascal's triangle
            l[i][j] = (l[i - 1][j - 1] + l[i - 1][j])

# Function to return the value of nCr
def nCr(n, r):
    # Return nCr
    return l[n][r]

# Driver code
# Build the Pascal's triangle
initialize()
n = 8
r = 3
print(nCr(n, r))

Javascript

<script>

// JavaScript implementation of the approach

// Initialize the matrix with 0
let l =
new Array(1001).fill(0).map(() => new Array(1001).fill(0));

function initialize()
{

    // 0C0 = 1
    l[0][0] = 1;
    for (let i = 1; i < 1001; i++) {
        // Set every nCr = 1 where r = 0
        l[i][0] = 1;
        for (let j = 1; j < i + 1; j++) {

            // Value for the current cell
            // of Pascal's triangle
            l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]);
        }
    }
}

// Function to return the value of nCr
function nCr(n, r)
{
    // Return nCr
    return l[n][r];
}

// Driver code

    // Build the Pascal's triangle
    initialize();
    let n = 8;
    let r = 3;
    document.write(nCr(n, r));

// This code is contributed by Mayank Tyagi

</script>

C#

// C# implementation of the approach

using System;
class GFG {
    // Initialize the matrix with 0
    static int[, ] l = new int[1001, 1001];

    static void initialize()
    {

        // 0C0 = 1
        l[0, 0] = 1;
        for (int i = 1; i < 1001; i++) {
            // Set every nCr = 1 where r = 0
            l[i, 0] = 1;
            for (int j = 1; j < i + 1; j++) {

                // Value for the current cell of Pascal's triangle
                l[i, j] = (l[i - 1, j - 1] + l[i - 1, j]);
            }
        }
    }

    // Function to return the value of nCr
    static int nCr(int n, int r)
    {
        // Return nCr
        return l[n, r];
    }
    // Driver code
    public static void Main()
    {
        // Build the Pascal's triangle
        initialize();
        int n = 8;
        int r = 3;
        Console.WriteLine(nCr(n, r));
    }
}

// This code is contributed by ihritik

Output

56

Complexity Analysis:

• Time Complexity: O(1). 
  The value of all pairs are precomputed so the time to answer the query is O(1), though some time is taken for precomputation but theoretically the precomputation takes constant time.
• Space Complexity: O(1). 
  Constant space is required.

Approach: 

The approach is called “Pascal’s Triangle Method”. It involves constructing Pascal’s triangle and then using the value of the corresponding cell to find nCr. The advantage of this method is that it saves time on calculating factorials by reusing previously computed values.

Steps:

1. Construct Pascal’s triangle with n+1 rows and n+1 columns.
2. Traverse to the cell corresponding to the values of n and r.
3. The value of that cell is the value of nCr.
4. Return the value of nCr.

C++

#include <iostream>
#include <vector>

using namespace std;

int pascals_triangle(int n, int r) {
    vector<vector<int>> triangle(n+1, vector<int>(n+1, 0));
    for (int i = 0; i <= n; i++) {
        triangle[i][0] = 1;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            triangle[i][j] = triangle[i-1][j-1] + triangle[i-1][j];
        }
    }
    return triangle[n][r];
}

int main() {
    int n = 5;
    int r = 3;
    int nCr = pascals_triangle(n, r);
    cout << nCr << endl;
    return 0;
}

Python3

def pascals_triangle(n, r):
    triangle = [[0]*(n+1) for _ in range(n+1)]
    for i in range(n+1):
        triangle[i][0] = 1
    for i in range(1, n+1):
        for j in range(1, i+1):
            triangle[i][j] = triangle[i-1][j-1] + triangle[i-1][j]
    return triangle[n][r]

n = 5
r = 3
nCr = pascals_triangle(n, r)
print(nCr)

C#

using System;
using System.Collections.Generic;

class Program {
    static int pascals_triangle(int n, int r) {
        List<List<int>> triangle = new List<List<int>>();
        for (int i = 0; i <= n; i++) {
            triangle.Add(new List<int>(new int[n+1]));
            triangle[i][0] = 1;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= i; j++) {
                triangle[i][j] = triangle[i-1][j-1] + triangle[i-1][j];
            }
        }
        return triangle[n][r];
    }

    static void Main(string[] args) {
        int n = 5;
        int r = 3;
        int nCr = pascals_triangle(n, r);
        Console.WriteLine(nCr);
    }
}

Output

10

Time Complexity: O(n^2)
Auxiliary Space: O(n^2)