In how many ways can 7 identical erasers be distributed among 4 kids

(3 2 1 1) can be arranged in 4!/2! = 12 ways

Also both cases are not independent, so you can't multiply.

We can see that each child gets a positive integer number of erasers. So let’s first list the ways 4 positive integers can sum to 7 (assuming any integer in the list can’t be greater than the previous one).

3 + 2 + 1 + 1 = 7 and 2 + 2 + 2 + 1 = 7

So we have two sets {3, 2, 1, 1} and {2, 2, 2, 1}. However, take the set {3, 2, 1, 1}, for example, and let’s say the four children are A, B, C, and D. A having 3, B having 2, C having 1 and D having 1 eraser is different from A having 1, B having 3, C having 2 and D having 1 eraser. So we need to ask ourselves: How many ways can we permute the four numbers in each set? The answer is to use the formula for permutation of indistinguishable items. Since we have:

4!/2! = 24/2 = 12 ways to permute the 4 numbers in {3, 2, 1, 1} and

4!/3! = 24/6 = 4 ways to permute the 4 numbers in {2, 2, 2, 1},

there are a total of 12 + 4 = 16 ways to distribute 7 erasers among 4 children in such a way that each child gets at least one eraser but no child gets more than 3 erasers.

Answer: C

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