### Concepts Used:

Mathematics

### Difficulty Level:

Hard

### Problem Statement (Simplified):

In this problem, we have to find a total number of ways to form a team of size

`K`

from`X`

men and`Y`

women with at least 4 men and 1 woman in the team.

**See original problem statement here**

#### Test Case:

```
Input:
1
5 2 6
Output:
7
Explanation:
For example, We have to form a team of size 6(K), and there need to be at least 4 men and 1 woman out of 5(X) men and 2(Y) women. So we can form 2 teams and in total 7 ways, consisting Men and Women in the following ways :
4 Men + 2 Women ( Total 5 ways )
5 Men + 1 Women ( Total 2 ways )
```

### Solving Approach :

1) We already know, we can select `m`

men and `n`

women from `X`

men and women in

ways, where any value

can be calculated using this formula :

2) As we know, we need to form a team of size `K`

where there need to be at least 4 men and 1 woman at least.

3) For the above condition, there will be `K-4`

different teams, containing different numbers of men and women in the team.

In the first team, there would be 4 men and `K-4`

women, in second-team there would be 5 men and `K-5`

women, as we reach to end, the last team consists of `K-1`

men and 1 woman.

4) So, mathematically the total number of the team becomes :

Total number of teams =

### Example:

Let’s assume, We have to form a team of size `6(K)`

and there needs to be atleast `4`

men and `1`

women out of `5(X)`

men and `2(Y)`

women. So we can form `2`

teams, consisting Men and Women in following ways :

`4`

Men + `2`

Women ( Total `5`

ways )

`5`

Men + `1`

Women ( Total `2`

ways )

*Total number of ways to form a team of size 6 with*`4`

*Men and*`2`

*women :*

Assuming Men(`M¹ M² M³ M⁴ M⁵`

) and Women (`W¹ M²`

) are there to select, now we have to select team of `6`

, we need `2`

women, so we’ll pick all women, and we need `4`

men, so we’ll pick `4`

men and then pair them with women. So possible ways are :

`M¹ M² M³ M⁴`

+`W¹ W²`

`M¹ M² M³ M⁵`

+`W¹ W²`

`M¹ M² M⁵ M⁴`

+`W¹ W²`

`M¹ M⁵ M³ M⁴`

+`W¹ W²`

`M⁵ M² M³ M⁴`

+`W¹ W²`

*Total number of ways to form a team of size 6 with*`5`

*Men and*`1`

*women :*

Assuming Men(`M¹ M² M³ M⁴ M⁵`

) and Women (`W¹ M²`

) are there to select, now we have to select team of `6`

, we need `5`

men, so we’ll pick all men, and we need `1`

woman, so we’ll pick `1`

woman and then pair her with men. So possible ways are :

`M¹ M² M³ M⁴ M⁴`

+ `W¹`

`M¹ M² M³ M⁴ M⁴`

+ `W¹`

Hence, in similar ways, we can find answers for different cases.

### Solutions:

#include <stdio.h> long long nCr(int n, int r){ long long value=1; for(int i=0;i<r;i++){ value*=(n-i); value/=(i+1); } return value; } int main() { int test; scanf("%d",&test); while(test--){ int n, m, k; scanf("%d%d%d",&n,&m,&k); long long sum=0; for(int i=4; i<k; i++){ sum += nCr(n,i)*nCr(m,k-i); } printf("%lld\n",sum); } }

#include <bits/stdc++.h> using namespace std; long long nCr(int n, int r){ long long value=1; for(int i=0;i<r;i++){ value*=(n-i); value/=(i+1); } return value; } int main() { int test; cin>test; while(test--){ int n, m, k; cin>n>m>k; long long sum=0; for(int i=4; i<k; i++){ sum += nCr(n,i)*nCr(m,k-i); } cout<<sum<<endl; } }

import java.util.*; import java.io.*; public class Main { static long nCr(int n, int r){ long value=1; for(int i=0;i<r;i++){ value*=(n-i); value/=(i+1); } return value; } public static void main(String args[]) throws IOException { Scanner sc = new Scanner(System.in); int test = sc.nextInt(); while(test--!=0){ int n = sc.nextInt(), m = sc.nextInt(), k = sc.nextInt(); long sum=0; for(int i=4; i<k; i++){ sum += nCr(n,i)*nCr(m,k-i); } System.out.println(sum); } } }

[forminator_quiz id="858"]

This article tried to discuss Mathematics. Hope this blog helps you understand and solve the problem. To practice more problems on Mathematics you can check out MYCODE | Competitive Programming.