In how many ways can $7$ identical erasers be distributed among $4$ kids in such a way that each kid gets at least one eraser but nobody gets more than $3$ erasers?

We need to distribute 7 identical erasers among 4 kids with two constraints:

Each kid must get at least 1 eraser and no kid can get more than 3 erasers.

Since the erasers are identical, we only care about how many each kid gets, not which specific erasers they receive.

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We give 1 eraser to each kid to satisfy the "at least 1" requirement.

Erasers used: 4 × 1 = 4 erasers

Erasers remaining: 7 - 4 = 3 erasers

Now each kid has 1 eraser, and we need to distribute the remaining 3 erasers such that no kid ends up with more than 3 erasers total.

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Since each kid already has 1 eraser, and the maximum allowed is 3 erasers per kid, each kid can receive at most 2 additional erasers.

We need to find all ways to distribute 3 identical items among 4 people where no person gets more than 2 items.

The possible distributions of these 3 remaining erasers are:

(2, 1, 0, 0): Give 2 extra to one kid, 1 extra to another kid

(1, 1, 1, 0): Give 1 extra to each of three kids

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Case 1: Distribution (2, 1, 0, 0)

We need to choose:

Which kid gets 2 extra erasers: 4 choices

Which kid gets 1 extra eraser: 3 remaining choices

Total ways: 4 × 3 = 12 ways

This is the same as ${{}}^{{4}}P_{{2}} = \dfrac{4!}{(4-2)!} = \dfrac{4!}{2!} = 12$

Case 2: Distribution (1, 1, 1, 0)

We need to choose which 3 kids out of 4 get 1 extra eraser each.

Total ways: ${{}}^{{4}}C_{{3}} = \dfrac{4!}{3!(4-3)!} = \dfrac{4!}{3! \times 1!} = 4$

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Total number of ways = Case 1 + Case 2 = 12 + 4 = 16 ways

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This approach is efficient because:

We first satisfy the minimum constraint by giving 1 eraser to each kid

We then focus on the remaining distribution while respecting the maximum constraint

We count systematically by considering all possible patterns

The key insight is that after satisfying the minimum requirement, we are left with a simpler problem of distributing fewer items with a clear upper bound.