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HMMT February

Combinatorics Difficulty 4.9 AIME Find the answer

Compute ∑n60=02∑n59=0n60⋯∑n2=0n3∑n1=0n2∑n0=0n11\sum_{n_{60}=0}^{2} \sum_{n_{59}=0}^{n_{60}} \cdots \sum_{n_{2}=0}^{n_{3}} \sum_{n_{1}=0}^{n_{2}} \sum_{n_{0}=0}^{n_{1}} 1

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The given sum counts the number of non-decreasing 61-tuples of integers (n0,…,n60)\left(n_{0}, \ldots, n_{60}\right) from the set {0,1,2}\{0,1,2\}. Such 61-tuples are in one-to-one correspondence with strictly increasing 61-tuples of integers (m0,…,m60)\left(m_{0}, \ldots, m_{60}\right) from the set {0,1,2,…,62}\{0,1,2, \ldots, 62\}: simply let mk=nk+km_{k}=n_{k}+k. But the number of such (m0,…,m60)\left(m_{0}, \ldots, m_{60}\right) is almost by definition (6361)=(632)=1953\binom{63}{61}=\binom{63}{2}=1953.

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