Six boxes labeled 1,2,…,6 are arranged in a line. Seven identical balls are to be placed into the boxes such that for any 1 ≤ k ≤ 6, there are at least k total balls amongst boxes 1,2,…k. How many different placements of the balls into the boxes are possible?
To solve this problem, we can use recursion and dynamic programming to count the number of valid placements. Let's define a function f(n, k) as the number of valid placements of n balls into k boxes.
The base cases are:
- If n = 0, then there is only one valid placement: all boxes are empty. So, f(0, k) = 1
- If k = 0 and n > 0, then there are no valid placements, so f(n, 0) = 0
For the recursive case, we can consider two possibilities:
1. If the last box is empty, then we can place the remaining n balls in the first k - 1 boxes, so f(n, k) = f(n, k-1).
2. If the last box is not empty, then we can place one ball in the last box (leaving n-1 balls) and distribute the remaining balls among the first k - 1 boxes. So, f(n, k) = f(n-1, k-1).
Using these recursive formulas and memoization to avoid redundant calculations, we can calculate the number of valid placements for any given n and k.
Let's calculate f(7, 6) using dynamic programming:
- f(0, k) = 1 for all k (base case)
- f(n, 0) = 0 for all n > 0 (base case)
Using the recursive formulas:
- f(1, 1) = f(0, 0) = 1
- f(2, 2) = f(1, 1) + f(1, 1) = 1 + 1 = 2
- f(3, 3) = f(2, 2) + f(2, 2) = 2 + 2 = 4
- f(4, 4) = f(3, 3) + f(3, 3) = 4 + 4 = 8
- f(5, 5) = f(4, 4) + f(4, 4) = 8 + 8 = 16
- f(6, 6) = f(5, 5) + f(5, 5) = 16 + 16 = 32
Therefore, there are 32 different placements of the 7 balls into the 6 labeled boxes.