We denote by $_6$ the number of sequences with 6 letters from the set {, , , }, where appears an even number of times, and appears an odd number of times.
(a) (True / False): $_6$ is the coefficient of $^6$ in the expression
$ (1 + ^2 + ^4 + ^6 + ⋯ )( + ^3 + ^5 + ⋯ )(1 + + ^2 + ^3 + ⋯ )^2 $
(b) (True / False): $_6$ is the coefficient of $\frac{^6}{6!}$ in the expression
$(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!}+ \cdots)\cdot (x + \frac{x^3}{3!} + \frac{x^5}{5!}+ \cdots)\cdot \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+ \cdots \right)^2$
(c) (True / False): $_6$ is the coefficient of $\frac{^6}{6!}$ in the expression
$(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!})\cdot (x + \frac{x^3}{3!} + \frac{x^5}{5!})\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!} + \frac{x^5}{5!} +\frac{x^6}{6!}\right)^2$
(d) (True / False): $_6$ is the coefficient of $\frac{^6}{6!}$ in the expression
$(1 + \frac{x^2}{2!} + \frac{x^4}{4!})\cdot (x + \frac{x^3}{3!} + \frac{x^5}{5!}+\frac{x^7}{7!})\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!} + \frac{x^5}{5!} +\frac{x^6}{6!}\right)^2$
So (a) is wrong because we have distinct letters where we don't care about the order, so we have an exponential generating function. But where it gets tricky is (d), because I believe that we get, for the first e.g.f., the terms until $x^4/4!$, because we can have a maximum of 4 A's in order to have an odd number of B's and the term $x^7/7!$ in the second e.g.f. is irrelevant for counting the number of sequences with 6 letters. But if my assumption is wrong and (d) is False then (b) and (c) is correct. Can somebody tell me which is the correct answer?