Numerical answer
In how many ways can 10 identical chocolates be distributed among 4 children so that every child gets at least 1 chocolate and no child gets more than 4?
Show answer and solution
Answer: 44
Give each child 1 first: we now distribute 6 more with each child getting 0 to 3 extra.
Without the upper limit: C(6 + 3, 3) = C(9, 3) = 84.
Subtract cases where some child gets ≥ 4 extra: give that child 4, distribute the remaining 2 freely: C(2 + 3, 3) = 10 ways, for each of 4 children ⇒ 40. Two children cannot both get ≥ 4 extra (that needs 8 > 6).
Answer = 84 − 40 = 44.
Tip: inclusion–exclusion on the upper bound is standard; always check whether two violations can occur together.
Without the upper limit: C(6 + 3, 3) = C(9, 3) = 84.
Subtract cases where some child gets ≥ 4 extra: give that child 4, distribute the remaining 2 freely: C(2 + 3, 3) = 10 ways, for each of 4 children ⇒ 40. Two children cannot both get ≥ 4 extra (that needs 8 > 6).
Answer = 84 − 40 = 44.
Tip: inclusion–exclusion on the upper bound is standard; always check whether two violations can occur together.
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