I need to find a combinatorial proof of this identity


I think inclusion exclusion is the best method here. But I"m having a really hard time coming up with a set to count. Permutations don't work here.

A hint would be really nice. Thanks.

  • $\begingroup$ $C_{2p}$ would be the $2p$th catalan number? $\endgroup$ Commented Nov 8, 2013 at 2:04
  • $\begingroup$ @BrianM.Scott Yes, as it was incorrect it is fixed now. $\endgroup$ Commented Nov 9, 2013 at 22:47

3 Answers 3


You have $n$ different pairs of shoes. How many ways are there to choose $n-1$ shoes so that you get at least one shoe from each pair? Obviously the answer is $0$, but you can also use inclusion-exclusion to count it the hard way as follows.

There are $\binom{2n}{n-1}$ ways to choose $n-1$ shoes. From this we want to exclude the ways that don’t include a shoe from the first pair; clearly there are $\binom{2n-2}{n-1}$ of those. We want to do the same for each of the $n$ pairs, so we should subtract $n\binom{2n-2}{n-1}$. But now each $(n-1)$-set that contains no shoe from either of the first two pairs has been counted once in $\binom{2n}{n-1}$ and subtracted twice in $n\binom{2n-2}{n-1}$ and should therefore be added back in. There are $\binom{2n-4}{n-1}$ such $(n-1)$-sets, and the same is true for each pair of the $\binom{n}2$ pairs of pairs, so we should add $\binom{n}2\binom{2n-4}{n-1}$.

Continuing in this fashion, we see that there are


sets of $n-1$ shoes that meet the requirements, so



Here is a proof using generating functions, quite a nice exercise in manipulation thereof. Rewrite the sum as follows: $$\sum_{k=0}^n {n\choose k} (-1)^k {2(n-k)\choose n-1}.$$

Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$ i.e. the product of the two generating functions is the generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$

Now clearly we have in the present case that $$A(z) = \sum_{q\ge 0} (-1)^q \frac{z^q}{q!} = \exp(-z).$$ For $B(z)$ we get that $$B_n(z) = \sum_{q\ge 0} {2q\choose n-1} \frac{z^q}{q!}$$ where $B_1(z) = \exp(z).$

In order to investigate the $B_n(z)$ we introduce the bivariate generating function $$Q(z,w) = \sum_{n\ge 1} B_n(z) w^n = \sum_{n\ge 1} w^n \sum_{q\ge 0} {2q\choose n-1} \frac{z^q}{q!} = \sum_{q\ge 0} \frac{z^q}{q!} \sum_{n\ge 1} w^n {2q\choose n-1} \\ = w \sum_{q\ge 0} \frac{z^q}{q!} \sum_{n\ge 1} {2q\choose n-1} w^{n-1} = w \sum_{q\ge 0} \frac{z^q}{q!} (w+1)^{2q} \\ = w \exp(z (w+1)^2) = \exp(z) \times w \times \exp(z (w^2+2w)).$$ Extracting coefficients from this we find that $$B_n(z) = [w^n] Q(z, w) = \exp(z) [w^{n-1}] \exp(z (w^2+2w)) \\= \exp(z) [w^{n-1}] \exp(z w^2) \exp(2 z w) = \exp(z) \sum_{a+2b=n-1} \frac{2^a z^a}{a!} \frac{z^b}{b!}.$$ This yields the formula $$B_n(z) = \exp(z) \sum_{b=0}^{\lfloor (n-1)/2 \rfloor} \frac{2^{n-1-2b}}{(n-1-2b)! \times b!} z^{n-1-b}.$$

The value of our sum is therefore $$n! [z^n] A(z) B_n(z) = n! [z^n] \sum_{b=0}^{\lfloor (n-1)/2 \rfloor} . \frac{2^{n-1-2b}}{(n-1-2b)! \times b!} z^{n-1-b} = 0$$ i.e. zero because as is easily seen the degree of the polynomial term is $n-1 < n.$

There is another calculation of this type at this MSE link -- I and at this MSE link -- II.


Let $\{S_{i}\}_{i=1}^{n}$ be a collection of sets each of size $2$. Double count sets $T$ of size $n-1$ from $\cup S_{i}$ such that $T$ contains an element from each $S_{i}$. THe RHS is clear. There are $n$ sets and $T$ has $n-1$ things this can never happen. The LHS use PIE.

  • $\begingroup$ What is PIE please ? $\endgroup$ Commented Nov 12, 2013 at 19:24
  • $\begingroup$ @EwanDelanoy Principal of inclusion exclusion. Sorry $\endgroup$ Commented Nov 12, 2013 at 19:25

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