Count ways choosing K balls from any given A boxes

Given integers A and K, there are A boxes the first box contains K balls, the Second box contains K + 1 balls, the Third Box contains K + 3 balls so on till the A'th box contains K + A balls, the task for this problem is to count the number of ways of picking K balls from any of the boxes. (Print answer modulo 10⁹ + 7).

Examples:

Input: A = 2, K = 2
Output: 4
Explanation: We have two boxes one with 2 balls let's say Box 1 with Balls {A1, A2} and one with 3 balls let's say Box 2 with balls {B1, B2, B3}

• First way: we can remove balls A1 and A2 from Box 1.
• Second way: we can remove balls B1 and B2 from Box 2.
• Third way: we can remove balls B2 and B3 from Box 2.
• Forth way : we can remove balls B1 and B3 from Box 2.

Input: A = 3, K = 3
Output: 15

Naive approach: The basic way to solve the problem is as follows:

The basic way to solve this problem is to generate all possible combinations by using a recursive approach.

Time Complexity: O(N!)
Auxiliary Space: O(1)

Efficient Approach:  The above approach can be optimized based on the following idea:

The problem can be solved with combinatorics:

• Total ways = ^KC_K + ^K+1C_K + ^K+2C_K + ....... +  ^K+A-1C_K
• we know ^N+1C_M+1 = ^NC_M + ^NC_M+1 that is ^NC_M = ^N+1C_M+1 - ^NC_M+1
• Total ways = ^K+1C_K+1 + (^K+2C_K+1 - ^K+1C_K+1) + (^K+3C_K+1 - ^K+2C_K+1) + ................ + (^K+AC_K+1 - ^K+A-1C_K-1)
• Total ways = ^K+AC_K+1
• Total ways = (K + A)! / ((K + 1)! * (A - 1)!)

Instead of dividing factorials of K + 1 and A - 1, we multiply their modular multiplicative inverses. 

Follow the steps below to solve the problem:

• Initializing fact[] array and Precomputing all factorials from 1 to 100000.
• initializing ANS variable.
• Calculating the answer by using the above formula.
• Print the answer.

Below is the implementation of the above approach:

4
15

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

Related Articles:

• Combinatorics Basics
• Modular multiplicative inverse