JEE Advanced 2024 Paper 2 · Mathematics · Permutations and Combinations · hard

Ten identical sweets are to be distributed among five children, each child receiving at most 3 sweets (a child may receive none). The number of ways…

Numerical answer

Ten identical sweets are to be distributed among five children, each child receiving at most

3

sweets (a child may receive none). The number of ways to do this is

Show answer and solution

Answer: 101

The count equals the coefficient of

x^{10}

(1+x+x^2+x^3)^5=\dfrac{(1-x^4)^5}{(1-x)^5}

. Expand:

(1-x^4)^5=1-5x^4+10x^8-\dots

and

(1-x)^{-5}=\sum\binom{n+4}{4}x^n

. The coefficient of

x^{10}

\binom{14}{4}-5\binom{10}{4}+10\binom{6}{4}=1001-1050+150=101

. Equivalently, by inclusion-exclusion: all distributions

\binom{14}{4}=1001

, minus those where some child gets at least

4

(

5\binom{10}{4}=1050

), plus those where two children each get at least

4

(

\binom52\binom{6}{4}=150

); three children would need at least

12

sweets, impossible. The answer is

101

, matching direct polynomial expansion.

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