A combinatorial identity

While tutoring today a student asked me about the following identity

(*)\qquad\qquad\qquad\displaystyle\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk=4^n.

One way to prove it is to count the number of possible triples (A,B,C) of sets with C\subseteq B\subseteq A\subseteq S=\{1,\dots,n\}. This can be done in two ways. First, if A, B, C contain i,j,k elements respectively then the number of such triples clearly equals the left-hand side of (*). On the other hand, since each element of S must belong to one of the four disjoint regions in the picture below

screenshot-2016-10-03-21

the number of such triples equals precisely 4^n.

An algebraic proof can be obtained by repeated use of the binomial theorem:

\displaystyle 4^n=\sum_{n\ge i\ge 0}\binom ni3^i=\sum_{n\ge i\ge j\ge 0}\binom ni\binom ij2^j=\sum_{n\ge i\ge j\ge k\ge 0}\binom ni\binom ij\binom jk.

Similarly, one obtains the more general identity

\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\binom{n}{i_1}\binom{i_1}{i_2}\cdots\binom{i_{m-1}}{i_m}=(m+1)^n.

This also follows from the multinomial theorem since the left-hand side equals

\displaystyle\sum_{n\ge i_m\ge\cdots\ge i_1\ge 0}\frac{n!}{(n-i_1)!(i_1-i_2)!\cdots (i_{m-1}-i_m)!i_m!}=(m+1)^n.

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