Count of Ways to Choose N People With at Least X Men and Y Women from P Men and Q Women | Set 2

Given integers N, P, Q, X, and Y, the task is to find the number of ways to form a group of N people having at least X men and Y women from P men and Q women, where (X + Y ≤ N, X ≤ P and Y ≤ Q).

Examples:

Input: P = 4, Q = 2, N = 5, X = 3, Y = 1
Output: 6
Explanation: Suppose given pool is {m1, m2, m3, m4} and {w1, w2}. Then possible combinations are:
m1 m2 m3 m4 w1
m1 m2 m3 m4 w2
m1 m2 m3 w1 w2
m1 m2 m4 w1 w2
m1 m3 m4 w1 w2
m2 m3 m4 w1 w2
Hence the count is 6.

Input: P = 5, Q = 2, N = 6, X = 4, Y = 1
Output: 7

Naive Approach: This problem is based on combinatorics, and details of the Naive approach is already discussed in Set-1 of this problem. 

For some general value of P, Q, N, X and Y we can calculate the total possible ways using the following formula:

where 

In this approach at every step we were calculating the value for each possible way. 

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: To solve this problem efficiently, we can use the Pascal Triangle property to calculate the , i.e.

1
1 1
1 2 1
1 3 3 1
.
.
.

which is nothing but

.
.
.

Follow the steps mentioned below:

• Use the pascal triangle to precalculate the values of the combination.
• Start iterating a loop from i = X to i = P and do the following for each iteration.
• Check if (N-i) ≥ Y and (N-i) ≤ Q.
• If the condition is satisfied then count the possible ways for i men and (N-i) women, otherwise, skip the step.
• Add the count with the total number of ways.
• Return the total count as your answer.

Below is the implementation of the approach:

C++

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

long long int pascal[31][31];

// Function to calculate the pascal triangle
void pascalTriangle()
{
    pascal[0][0] = 1;
    pascal[1][0] = 1;
    pascal[1][1] = 1;

    // Loop to calculate values of
    // pascal triangle
    for (int i = 2; i < 31; i++) {
        pascal[i][0] = 1;
        for (int j = 1; j < i; j++)
            pascal[i][j]
                = pascal[i - 1][j]
                  + pascal[i - 1][j - 1];
        pascal[i][i] = 1;
    }
}

// Function to calculate the number of ways
long long int countWays(int n, int p,
                        int q, int x,
                        int y)
{

    // Variable to store the answer
    long long int sum = 0;

    // Loop to calculate the number of ways
    for (long long int i = x; i <= p; i++) {
        if (n - i >= y && n - i <= q)
            sum += pascal[p][i]
                   * pascal[q][n - i];
    }
    return sum;
}

// Driver code
int main()
{
    pascalTriangle();

    int P = 4, Q = 2, N = 5, X = 3, Y = 1;

    // Calculate possible ways for given
    // N, P, Q, X and Y
    cout << countWays(N, P, Q, X, Y)
         << endl;
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
class GFG {

    static long pascal[][] = new long[31][31];

    // Function to calculate the pascal triangle
    static void pascalTriangle()
    {
        pascal[0][0] = 1;
        pascal[1][0] = 1;
        pascal[1][1] = 1;

        // Loop to calculate values of
        // pascal triangle
        for (int i = 2; i < 31; i++) {
            pascal[i][0] = 1;
            for (int j = 1; j < i; j++)
                pascal[i][j] = pascal[i - 1][j]
                               + pascal[i - 1][j - 1];
            pascal[i][i] = 1;
        }
    }

    // Function to calculate the number of ways
    static long countWays(int n, int p, int q, int x, int y)
    {

        // Variable to store the answer
        long sum = 0;

        // Loop to calculate the number of ways
        for (int i = x; i <= p; i++) {
            if (n - i >= y && n - i <= q)
                sum += pascal[p][i] * pascal[q][n - i];
        }
        return sum;
    }

    // Driver code
    public static void main(String[] args)
    {
        pascalTriangle();
        int P = 4, Q = 2, N = 5, X = 3, Y = 1;

        // Calculate possible ways for given
        // N, P, Q, X and Y
        System.out.println(countWays(N, P, Q, X, Y));
    }
}

// This code is contributed by Potta Lokesh

Python3

pascal = [[0 for i in range(31)] for j in range(31)]

# Function to calculate the pascal triangle
def pascalTriangle():
  pascal[0][0] = 1;
  pascal[1][0] = 1;
  pascal[1][1] = 1;

  # Loop to calculate values of
  # pascal triangle
  for i in range(2, 31):
    pascal[i][0] = 1;
    for j in range(i):
      pascal[i][j] = pascal[i - 1][j] + pascal[i - 1][j - 1];
    pascal[i][i] = 1;

# Function to calculate the number of ways
def countWays(n, p, q, x, y):

  # Variable to store the answer
  sum = 0;

  # Loop to calculate the number of ways
  for i in range(x, p + 1):
    if (n - i >= y and n - i <= q):
      sum += pascal[p][i] * pascal[q][n - i];
  return sum;

# Driver code
pascalTriangle();

P = 4
Q = 2
N = 5
X = 3
Y = 1;

# Calculate possible ways for given
# N, P, Q, X and Y
print(countWays(N, P, Q, X, Y))

# This code is contributed by Saurabh Jaiswal

C#

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

class GFG{

static long [,]pascal = new long[31, 31];

// Function to calculate the pascal triangle
static void pascalTriangle()
{
    pascal[0, 0] = 1;
    pascal[1, 0] = 1;
    pascal[1, 1] = 1;

    // Loop to calculate values of
    // pascal triangle
    for(int i = 2; i < 31; i++)
    {
        pascal[i, 0] = 1;
        for(int j = 1; j < i; j++)
            pascal[i, j] = pascal[i - 1, j] +
                           pascal[i - 1, j - 1];
        pascal[i, i] = 1;
    }
}

// Function to calculate the number of ways
static long countWays(int n, int p, int q, int x, int y)
{

    // Variable to store the answer
    long sum = 0;

    // Loop to calculate the number of ways
    for(int i = x; i <= p; i++)
    {
        if (n - i >= y && n - i <= q)
            sum += pascal[p, i] * pascal[q, n - i];
    }
    return sum;
}

// Driver code
public static void Main(String[] args)
{
    pascalTriangle();

    int P = 4, Q = 2, N = 5, X = 3, Y = 1;

    // Calculate possible ways for given
    // N, P, Q, X and Y
    Console.WriteLine(countWays(N, P, Q, X, Y));
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>
let pascal = new Array(31).fill(0).map(() => new Array(31).fill(0));

// Function to calculate the pascal triangle
function pascalTriangle() {
  pascal[0][0] = 1;
  pascal[1][0] = 1;
  pascal[1][1] = 1;

  // Loop to calculate values of
  // pascal triangle
  for (let i = 2; i < 31; i++) {
    pascal[i][0] = 1;
    for (let j = 1; j < i; j++)
      pascal[i][j]
        = pascal[i - 1][j]
        + pascal[i - 1][j - 1];
    pascal[i][i] = 1;
  }
}

// Function to calculate the number of ways
function countWays(n, p, q, x, y) {

  // Variable to store the answer
  let sum = 0;

  // Loop to calculate the number of ways
  for (let i = x; i <= p; i++) {
    if (n - i >= y && n - i <= q)
      sum += pascal[p][i]
        * pascal[q][n - i];
  }
  return sum;
}

// Driver code
pascalTriangle();

let P = 4, Q = 2, N = 5, X = 3, Y = 1;

// Calculate possible ways for given
// N, P, Q, X and Y
document.write(countWays(N, P, Q, X, Y))

// This code is contributed by gfgking.
</script>

Output

6

Time Complexity: O(N)
Auxiliary Space: O(N²)