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 = 7w = 6n = 5k = 3print(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.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
