Combinatorics problem using generating functions 1 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!
 A: 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.
