Given four integers n, w, m and k where,
- m is the total number of men.
- w is the total number of women.
- n is the total number of people that need to be selected to form the team.
- k is the minimum number of men that have to be selected.
The task is to find the number of ways in which the team can be formed.
Examples:
Input: m = 2, w = 2, n = 3, k = 1
Output: 4
There are 2 men, 2 women. We need to make a team of size 3 with at least one man and one woman. We can make the team in following ways.
m1 m2 w1
m1 w1 w2
m2 w1 w2
m1 m2 w2Input: m = 7, w = 6, n = 5, k = 3
Output: 756
Input: m = 5, w = 6, n = 6, k = 3
Output: 281
Approach: Since, we have to take at least k men.
Totals ways = Ways when ‘k’ men are selected + Ways when ‘k+1’ men are selected + … + when ‘n’ men are selected
.
Taking the first example from above where out of 7 men and 6 women, total 5 people need to be selected with at least 3 men,
Number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
= 7 x 6 x 5 x 6 x 5 + (7C3 x 6C1) + (7C2)
= 525 + 7 x 6 x 5 x 6 + 7 x 6
= (525 + 210 + 21)
= 756
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Returns factorial // of the number int fact( int n) { int fact = 1; for ( int i = 2; i <= n; i++) fact *= i; return fact; } // Function to calculate ncr int ncr( int n, int r) { int ncr = fact(n) / (fact(r) * fact(n - r)); return ncr; } // Function to calculate // the total possible ways int ways( int m, int w, int n, int k) { int ans = 0; while (m >= k) { ans += ncr(m, k) * ncr(w, n - k); k += 1; } return ans; } // Driver code int main() { int m, w, n, k; m = 7; w = 6; n = 5; k = 3; cout << ways(m, w, n, k); } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Returns factorial // of the number static int fact( int n) { int fact = 1 ; for ( int i = 2 ; i <= n; i++) fact *= i; return fact; } // Function to calculate ncr static int ncr( int n, int r) { int ncr = fact(n) / (fact(r) * fact(n - r)); return ncr; } // Function to calculate // the total possible ways static int ways( int m, int w, int n, int k) { int ans = 0 ; while (m >= k) { ans += ncr(m, k) * ncr(w, n - k); k += 1 ; } return ans; } // Driver code public static void main (String[] args) { int m, w, n, k; m = 7 ; w = 6 ; n = 5 ; k = 3 ; System.out.println( ways(m, w, n, k)); } } // This Code is contributed // by shs |
Python3
# Python 3 implementation of the approach # Returns factorial of the number def fact(n): fact = 1 for i in range ( 2 , n + 1 ): fact * = i return fact # Function to calculate ncr def ncr(n, r): ncr = fact(n) / / (fact(r) * fact(n - r)) return ncr # Function to calculate # the total possible ways def ways(m, w, n, k): ans = 0 while (m > = k): ans + = ncr(m, k) * ncr(w, n - k) k + = 1 return ans; # Driver code m = 7 w = 6 n = 5 k = 3 print (ways(m, w, n, k)) # This code is contributed by sahishelangia |
C#
// C# implementation of the approach class GFG { // Returns factorial // of the number static int fact( int n) { int fact = 1; for ( int i = 2; i <= n; i++) fact *= i; return fact; } // Function to calculate ncr static int ncr( int n, int r) { int ncr = fact(n) / (fact(r) * fact(n - r)); return ncr; } // Function to calculate // the total possible ways static int ways( int m, int w, int n, int k) { int ans = 0; while (m >= k) { ans += ncr(m, k) * ncr(w, n - k); k += 1; } return ans; } // Driver code static void Main () { int m, w, n, k; m = 7; w = 6; n = 5; k = 3; System.Console.WriteLine( ways(m, w, n, k)); } } // This Code is contributed by mits |
PHP
<?php // PHP implementation of the approach // Returns factorial of the number function fact( $n ) { $fact = 1; for ( $i = 2; $i <= $n ; $i ++) $fact *= $i ; return $fact ; } // Function to calculate ncr function ncr( $n , $r ) { $ncr = (int)(fact( $n ) / (fact( $r ) * fact( $n - $r ))); return $ncr ; } // Function to calculate the total // possible ways function ways( $m , $w , $n , $k ) { $ans = 0; while ( $m >= $k ) { $ans += ncr( $m , $k ) * ncr( $w , $n - $k ); $k += 1; } return $ans ; } // Driver code $m = 7; $w = 6; $n = 5; $k = 3; echo ways( $m , $w , $n , $k ); // This Code is contributed // by Mukul Singh |
756
Further Optimization : The above code can be optimized using faster algorithms for binomial coefficient computation.
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