In how many ways can you divide $n$ different balls into $5$ different boxes so that the two last boxes has an even number of balls.
I've been given a clue to show that: $\sum_{n=0}^\infty {x^{2n} \over (2n)!} = {e^x + e^{-x} \over 2}$
My thought was using the $e^x$ Taylor's series, but i don't really getting anything.
I'll be happy to get partial solution , or any explaination on how should I approach this problem.
Thanks in advance!
The combinatorial species here is
$$\mathfrak{S}_{=3}(\mathfrak{P}(\mathcal{Z})) \mathfrak{S}_{=2}(\mathfrak{P}_{\mathrm{even}}(\mathcal{Z})).$$
This immediately produces the generating function
$$G(z) = \exp(z)^3 \frac{(\exp(z)+\exp(-z))^2}{4}.$$
which is
$$G(z) = \frac{1}{4}(\exp(5z)+2\exp(3z)+\exp(z)).$$
Coefficient extraction produces the formula $$n! [z^n] G(z) = \frac{1}{4} 5^n + \frac{1}{2} 3^n + \frac{1}{4}.$$
This yields the sequence $$3, 11, 45, 197, 903, 4271, 20625, 100937, 498123, 2470931, 12295605,\ldots$$
which is OEIS A146086.
The OEIS confirms the above derivation because it says we are counting numbers consisting of $n$ odd digits where the one and three digit occur an even number of times. This is the same as partitioning the $n$ positions into three sets for the five digit, the seven digit and the nine digit and two sets for the one digit and the three digit, where the latter two must contain an even number of elements. This is precisely the problem definition.
Addendum. Consulting the linked to answer from the comments which I hadn't seen it appears to be correct and complete and hence renders this question a duplicate.