# Count the number of ways to fill K boxes with N distinct items

Given two values N and K. Find the number of ways to arrange the N distinct items in the boxes such that exactly K (K<N) boxes are used from the N distinct boxes. The answer can be very large so return the answer modulo 109 + 7.

Note: 1 <= N <= K <= 105.
Prerequisites: Factorial of a number, Compute nCr % p

Examples:

Input: N = 5, k = 5
Output: 120

Input: N = 5, k = 3
Output: 1500

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We will use the inclusion-exclusion principle to count the ways.

1. Let us assume that the boxes are numbered 1 to N and now we have to choose any K boxes and use them. The number of ways to do this is NCK.
2. Now any item can be put in any of the chosen boxes, hence the number of ways to arrange them is KN But here, we may count arrangements with some boxes empty. Hence, we will use the inclusion-exclusion principle to ensure that we count ways with all K boxes filled with at least one item.
3. Let us understand the application of the inclusion-exclusion principle:

• So out of KN ways, we subtract the case when at least 1 box(out of K) is empty. Hence, subtract
(KC1)*((K-1)N).
• Note that here, The cases where exactly two boxes are empty are subtracted twice(once when we choose the first element in (KC1) ways, and then when we choose the second element in (KC1) ways).
• Hence, we add these ways one time to compensate. So we add (KC2)*((K – 2)N).
• Similarly, here we need to add the number of ways when at least 3 boxes were empty, and so on…
4. Hence, the total number of ways: ## C++

 `// C++ program to calculate the  ` `// above formula ` `#include ` `#define mod 1000000007 ` `#define int long long ` ` `  `using` `namespace` `std; ` ` `  `// To store the factorials  ` `// of all numbers ` `int` `factorial; ` ` `  `// Function to calculate factorial  ` `// of all numbers ` `void` `StoreFactorials(``int` `n) ` `{ ` `    ``factorial = 1; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `    ``{ ` `        ``factorial[i] =  ` `          ``(i * factorial[i - 1]) ` `            ``% mod; ` `         `  `    ``} ` `} ` ` `  `// Calculate x to the power y  ` `// in O(log n) time ` `int` `Power(``int` `x, ``int` `y) ` `{ ` `    ``int` `ans = 1; ` `    ``while` `(y > 0) { ` `        ``if` `(y % 2 == 1) { ` `            ``ans = (ans * x) % mod; ` `        ``} ` `        ``x = (x * x) % mod; ` `        ``y /= 2; ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Function to find inverse mod of  ` `// a number x ` `int` `invmod(``int` `x) ` `{ ` `    ``return` `Power(x, mod - 2); ` `} ` ` `  `// Calculate (n C r) ` `int` `nCr(``int` `n, ``int` `r) ` `{ ` `    ``return` `(factorial[n]  ` `            ``* invmod((factorial[r] ` `            ``* factorial[n - r]) % mod)) ` `            ``% mod; ` `} ` ` `  `int` `CountWays(``int` `n,``int` `k) ` `{ ` `    ``StoreFactorials(n); ` ` `  `     `  `    ``// Loop to compute the formula  ` `    ``// evaluated ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = k; i >= 0; i--) ` `    ``{ ` `        ``if` `(i % 2 == k % 2)  ` `        ``{ ` `            ``// Add even power terms ` `            ``ans = (ans + (Power(i, n) ` `                  ``* nCr(k, i)) % mod)  ` `                  ``% mod; ` `        ``} ` `        ``else`  `        ``{ ` `            ``// Subtract odd power terms ` `            ``ans = (ans + mod - (Power(i, n)  ` `                  ``* nCr(k, i)) % mod) % mod; ` `        ``} ` `    ``} ` `     `  `    ``// Choose the k boxes which  ` `    ``// were used ` `    ``ans = (ans * nCr(n, k)) % mod; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `signed` `main() ` `{ ` `    ``int` `N = 5; ` `    ``int` `K = 5; ` `     `  `    ``cout << CountWays(N, K) << ``"\n"``; ` `     `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program to calculate the  ` `# above formula ` ` `  `mod ``=` `1000000007` ` `  `# To store the factorials  ` `# of all numbers ` `factorial ``=` `[``0` `for` `i ``in` `range``(``100005``)] ` ` `  `# Function to calculate factorial  ` `# of all numbers ` `def` `StoreFactorials(n): ` `     `  `    ``factorial[``0``] ``=` `1` `    ``for` `i ``in` `range``(``1``, n ``+` `1``, ``1``): ` `        ``factorial[i] ``=` `(i ``*` `factorial[i ``-` `1``]) ``%` `mod ` ` `  `# Calculate x to the power y  ` `# in O(log n) time ` `def` `Power(x, y): ` `     `  `    ``ans ``=` `1` `    ``while` `(y > ``0``): ` `         `  `        ``if` `(y ``%` `2` `=``=` `1``): ` `            ``ans ``=` `(ans ``*` `x) ``%` `mod ` `             `  `        ``x ``=` `(x ``*` `x) ``%` `mod ` `        ``y ``/``/``=` `2` `         `  `    ``return` `ans ` ` `  `# Function to find inverse mod  ` `# of a number x ` `def` `invmod(x): ` `     `  `    ``return` `Power(x, mod ``-` `2``) ` ` `  `# Calculate (n C r) ` `def` `nCr(n, r): ` `    ``return` `((factorial[n] ``*` `invmod((factorial[r] ``*`  `                                    ``factorial[n ``-` `r]) ``%`  `                                    ``mod)) ``%` `mod) ` ` `  `def` `CountWays(n, k): ` `     `  `    ``StoreFactorials(n) ` `     `  `    ``# Loop to compute the formula  ` `    ``# evaluated ` `    ``ans ``=` `0` `    ``i ``=` `k ` `     `  `    ``while``(i >``=` `0``): ` `        ``if` `(i ``%` `2` `=``=` `k ``%` `2``): ` `             `  `            ``# Add even power terms ` `            ``ans ``=` `((ans ``+` `(Power(i, n) ``*`  `                           ``nCr(k, i)) ``%` `mod) ``%` `mod) ` `        ``else``: ` `             `  `            ``# Subtract odd power terms ` `            ``ans ``=` `((ans ``+` `mod ``-` `(Power(i, n) ``*` `                                 ``nCr(k, i)) ``%`  `                                 ``mod) ``%` `mod) ` `        ``i ``-``=` `1` `         `  `    ``# Choose the k boxes which  ` `    ``# were used ` `    ``ans ``=` `(ans ``*` `nCr(n, k)) ``%` `mod ` `     `  `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``N ``=` `5` `    ``K ``=` `5` `     `  `    ``print``(CountWays(N, K)) ` ` `  `# This code is contributed by Surendra_Gangwar `

Output:

```120
```

Time complexity: O(N*log N)

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