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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 | Set 2

  • Last Updated : 06 Dec, 2021

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 girl 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

 

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 and N we can calculate the total possible ways using the following formula:

_{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)!}

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 _{r}^{n}\textrm{C}         , i.e.

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

which is nothing but

_{0}^{0}\textrm{C}
_{0}^{1}\textrm{C}           _{1}^{1}\textrm{C}
_{0}^{2}\textrm{C}           _{1}^{2}\textrm{C}           _{2}^{2}\textrm{C}
_{0}^{3}\textrm{C}           _{1}^{3}\textrm{C}           _{2}^{3}\textrm{C}           _{3}^{3}\textrm{C}
.
.
.

Follow the steps mentioned below:

  • Use the pascal triangle to precalculate the values of the combination.
  • Start iterating a loop from i = 4 to i = P and do the following for each iteration.
  • Check if (N-i) ≥ 1 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)
{
 
    // Variable to store the answer
    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 += pascal[p][i]
                   * pascal[q][n - i];
    }
    return sum;
}
 
// Driver code
int main()
{
    pascalTriangle();
 
    int P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    cout << countWays(N, P, Q) << endl;
    return 0;
}

Java




import java.util.*;
public 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] = (int)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)
{
 
    // Variable to store the answer
    long sum = 0;
 
    // Loop to calculate the number of ways
    for (int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q) {
            sum += (int)pascal[p][i]
                   * (int)pascal[q][n - i];
        }
    }
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    pascalTriangle();
 
    int P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    System.out.print(countWays(N, P, Q));
 
}
}
// This code is contributed by Samim Hossain Mondal.

Python3




# Python3 program for the above approach
import numpy as np
 
pascal = np.zeros((31,31));
 
# 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(1, 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) :
 
    # Variable to store the answer
    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 += pascal[p][i] * pascal[q][n - i];
 
    return int(sum);
 
# Driver code
if __name__ ==  "__main__" :
 
    pascalTriangle();
 
    P = 5; Q = 2; N = 5;
 
    # Calculate possible ways for given
    # N, P, and Q
    print(countWays(N, P, Q));
 
    # This code is contributed by AnkThon

C#




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] = (int)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)
{
 
    // Variable to store the answer
    long sum = 0;
 
    // Loop to calculate the number of ways
    for (int i = 4; i <= p; i++) {
        if (n - i >= 1 && n - i <= q) {
            sum += (int)pascal[p, i]
                   * (int)pascal[q, n - i];
        }
    }
    return sum;
}
 
// Driver code
public static void Main()
{
    pascalTriangle();
 
    int P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    Console.Write(countWays(N, P, Q));
 
}
}
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
 
    let pascal = new Array(31).fill(0).map(() => new Array(31).fill(0));
 
    // Function to calculate the pascal triangle
    const 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
    const countWays = (n, p, q) => {
 
        // Variable to store the answer
        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 += pascal[p][i]
                    * pascal[q][n - i];
        }
        return sum;
    }
 
    // Driver code
    pascalTriangle();
 
    let P = 5, Q = 2, N = 5;
 
    // Calculate possible ways for given
    // N, P, and Q
    document.write(countWays(N, P, Q));
 
// This code is contributed by rakeshsahni
 
</script>

 
 

Output
10

 

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

 


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