A combinatorial question about odd and even numbers Consider an $n$ tuple. Each index of the tuple is filled with a symbol chosen from the following set comprising of four symbols:
\begin{equation}
S = \{a, b, c, d\}.
\end{equation}
Note that there are $4^{n}$ possible fillings.
I want to count the number of possible tuples such that $b$-s, $c$-s, and $d$-s all occur in it an even number of times (not necessarily the same even number for each.)
For example, if $n = 4$, then $bbba$ is not a valid tuple --- as it has $3$ $b$-s, which is an odd number. However $bb cc$ is a valid tuple — as both $b$ and $c$ occur an even number of times.

For $n = 2$, the valid choices are $aa, bb, cc, dd$.
Hence, the number of such tuples is $4$.
For $n = 3$, the valid choices are $aaa, abb, acc, add, bba, cca, dda, bab, cac, dad$.
So, $10$ choices.
Is there a general pattern?

Since this is tagged combinatorial-proofs, here’s a purely bijective solution. Let the “signature” of a tuple be the numbers of $b$s, $c$s, and $d$s mod 2, and let $x_s$ be the number of tuples with signature $s$ (so we seek $x_{000}$).
By the symmetry $b → c → d → b$, we have $x_{100} = x_{010} = x_{001}$ and $x_{110} = x_{101} = x_{011}$. We can count all $4^n$ tuples as follows:
$$x_{000} + 3x_{100} + 3x_{011} + x_{111} = 4^n. \tag{1}$$
Consider the transformation that flips the first letter that’s $a$ or $b$ to $b$ or $a$, respectively. This is a self-inverse transformation that exchanges signatures $000 ↔ 100$, $010 ↔ 110$, $001 ↔ 101$, $011 ↔ 111$, except that it fails on the $2^n$ tuples made entirely from $c$ and $d$. If $n$ is even, the signatures of these failing tuples are half $000$ and half $011$; if $n$ is odd, they’re instead half $010$ and half $001$. So:
\begin{gather*}
  x_{000} - x_{100} = 2^{n - 2} + (-2)^{n - 2}, \tag{2} \\
  x_{011} - x_{111} = 2^{n - 2} + (-2)^{n - 2}, \tag{3} \\
  x_{010} - x_{110} = 2^{n - 2} - (-2)^{n - 2}, \tag{4} \\
  x_{001} - x_{101} = 2^{n - 2} - (-2)^{n - 2}. \tag{5} \\
  \end{gather*}
By $\text{(1)} + 7·\text{(2)} + \text{(3)} + 4·\text{(4)}$, we have
$$8x_{000} = 4^n + 8(2^{n - 2} + (-2)^{n - 2}) + 4(2^{n - 2} - (-2)^{n - 2}), \\
x_{000} = \frac{4^n + 3·2^n + (-2)^n}{8}.$$
