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Count of ways to choose N people containing at least 4 boys and 1 girl from P boys and Q girls

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  • Last Updated : 02 Feb, 2022

Given integers N, P, and Q the task is to find the number of ways to form a group of N people having at least 4 boys and 1 girls from P boys and Q girls.

Examples:

Input:  P = 5, Q = 2, N = 5
Output: 10
Explanation:  Suppose given pool is {m1, m2, m3, m4, m5} and {w1, w2}. Then possible combinations are: 

m1 m2 m3 m4 w1
m2 m3 m4 m5 w1
m1 m3 m4 m5 w1
m1 m2 m4 m5 w1
m1 m2 m3 m5 w1
m1 m2 m3 m4 w2
m2 m3 m4 m5 w2
m1 m3 m4 m5 w2
m1 m2 m4 m5 w2
m1 m2 m3 m5 w2

Hence the count is 10.

Input:  P = 5, Q = 2, N = 6
Output: 7

 

Approach: This problem is based on combinatorics where we need to select at least 4 boys out of 1 boys available, and at least Y girls out of Q girls available, so that total people selected are N.
Consider the example:

 P = 5, Q = 2, N = 6 
In this, the possible selections are:
(4 boys out of 5) * (2 girls out of 2) + (5 boys out of 5) * (1 girl out of 2)
= 5C4 * 2C2 + 5C5 * 2C1

So for some general values of P, Q and N, the approach can be visualised as:

_{4}^{P}\textrm{C} \ast _{N-4}^{Q}\textrm{C} + _{5}^{P}\textrm{C} \ast _{N-5}^{Q}\textrm{C} + . . . + _{N-2}^{P}\textrm{C} \ast _{2}^{Q}\textrm{C} + _{N-1}^{P}\textrm{C} \ast _{1}^{Q}\textrm{C}                
where  
_{r}^{n}\textrm{C} = \frac{n!}{r!*(n-r)!}

Follow the steps mentioned below to implementation it:

  • Start iterating a loop from i = 4 till i = P.
  • At each iteration calculate the number of possible ways if we choose i boys and (N-i) girls, using combination
     _{i}^{P}\textrm{C} \ast _{N-i}^{Q}\textrm{C}
  • Add the possible value for each iteration as the total number of ways.
  • Return the total calculated ways at the end.

Below is the implementation of the approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate factorial
long long int fact(int f)
{
    f++;
    long long int ans = 1;
 
    // Loop to calculate factorial of f
    while (--f > 0)
        ans = ans * f;
    return ans;
}
 
// Function to calculate combination nCr
long long int ncr(int n, int r)
{
    return (fact(n) / (fact(r) * fact(n - r)));
}
 
// Function to calculate the number of ways
long long int countWays(int n, int p, int q)
{
    long long int sum = 0;
 
    // Loop to calculate the number of ways
    for (long long int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q)
            sum += (ncr(p, i) * ncr(q, n - i));
    }
    return sum;
}
 
// Driver code
int main()
{
    int P = 5, Q = 2, N = 5;
    cout << countWays(N, P, Q) << endl;
    return 0;
}

Java




import java.util.*;
 
class GFG{
 
// Function to calculate factorial
static int fact(int f)
{
    f++;
    int ans = 1;
 
    // Loop to calculate factorial of f
    while (--f > 0)
        ans = ans * f;
    return ans;
}
 
// Function to calculate combination nCr
static int ncr(int n, int r)
{
    return (fact(n) / (fact(r) * fact(n - r)));
}
 
// Function to calculate the number of ways
static int countWays(int n, int p, int q)
{
    int sum = 0;
 
    // Loop to calculate the number of ways
    for (int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q)
            sum += (ncr(p, i) * ncr(q, n - i));
    }
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    int P = 5, Q = 2, N = 5;
    System.out.print(countWays(N, P, Q) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Function to calculate factorial
def fact(f):
    ans = 1
 
    # Loop to calculate factorial of f
    while (f):
        ans = ans * f
        f -= 1
    return ans
 
# Function to calculate combination nCr
def ncr(n, r):
    return (fact(n) / (fact(r) * fact(n - r)))
 
# Function to calculate the number of ways
def countWays(n, p, q):
    sum = 0
 
    # Loop to calculate the number of ways
    for i in range(4, p + 1):
        if (n - i >= 1 and n - i <= q):
            sum += (ncr(p, i) * ncr(q, n - i))
 
    return (int)(sum)
 
# Driver code
P = 5
Q = 2
N = 5
print(countWays(N, P, Q))
 
 # This code is contributed by gfgking.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
// Function to calculate factorial
static int fact(int f)
{
    f++;
    int ans = 1;
 
    // Loop to calculate factorial of f
    while (--f > 0)
        ans = ans * f;
    return ans;
}
 
// Function to calculate combination nCr
static int ncr(int n, int r)
{
    return (fact(n) / (fact(r) * fact(n - r)));
}
 
// Function to calculate the number of ways
static int countWays(int n, int p, int q)
{
    int sum = 0;
 
    // Loop to calculate the number of ways
    for (int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q)
            sum += (ncr(p, i) * ncr(q, n - i));
    }
    return sum;
}
 
    // Driver Code
    public static void Main()
    {
        int P = 5, Q = 2, N = 5;
        Console.Write(countWays(N, P, Q));
    }
}
 
// This code is contributed by sanjoy_62.

Javascript




<script>
 
    // Function to calculate factorial
    const fact = (f) => {
        f++;
        let ans = 1;
 
        // Loop to calculate factorial of f
        while (--f > 0)
            ans = ans * f;
        return ans;
    }
 
    // Function to calculate combination nCr
    const ncr = (n, r) => {
        return (fact(n) / (fact(r) * fact(n - r)));
    }
 
    // Function to calculate the number of ways
    const countWays = (n, p, q) => {
        let sum = 0;
 
        // Loop to calculate the number of ways
        for (let i = 4; i <= p; i++) {
            if (n - i >= 1 && n - i <= q)
                sum += (ncr(p, i) * ncr(q, n - i));
        }
        return sum;
    }
 
    // Driver code
 
    let P = 5, Q = 2, N = 5;
    document.write(countWays(N, P, Q));
 
    // This code is contributed by rakeshsahni
 
</script>
Output
10

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


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