This is a question that was asked by one of the members but was down-voted as he did not show his attempt to solve it. He added that later and the down-votes were taken back. I answered the question but for some reason, OP deleted the post quickly after that so I could not add any comments. As I was solving the problem, I found something interesting that I wanted some insights on and that is the reason, I am posting it with additional details.
Question: Person $1$ lists all possible strings of length $2x$, such that each letter is one of $A, B$, and each string containts the same number of $A's$ and $B's$.
Person $2$ lists all possible strings of length $x$, such that each letter is one of $A, B, C, D$ and each string contains the same number of $A's$ and $B's$ ($A = B = 0$ or $C = 0$, $D = 0$ are allowed).
My solution:
Person $1$:
Number of strings of length $2x$ from equal numbers of $A$ and $B$ ($x$ each) $ = M(2x) = \dfrac {(2x)!}{x!x!}$, ($x \in \mathbb {Z+}$)
Person $2$:
we can start from $A = B = 0$. Each alphabet in the string can have value of $C$ or $D$ independently.
$N(0) = 2^x$
For $A = B = 1, N(1) = {^x}P_2 \times 2^{x-2}$
For $A = B = 2, N(2) = \dfrac {{^x}P_4}{2!2!} \times 2^{x-4}$
For $A = B = k, 2k \le x, N(k) = \dfrac {{^x}P_{2k}}{k!k!} \times 2^{x-2k}$
Say, $n = \left\lfloor\dfrac{x}{2}\right\rfloor$, where $n$ is the largest integer lower than or equal to $\dfrac{x}{2}$.
$N(x) = \sum \limits_{k = 0}^{n}N(k) = \sum \limits_{k = 0}^{n} \dfrac {{^x}P_{2k}}{k!k!} \times 2^{x-2k}$
I checked and noticed $M(2x) = N(x)$ for $2 \le x \le 5$.
So the question I have, is it only happening for smaller values of $x$ or is the below true?
$\sum \limits_{k = 0}^{n} \dfrac {{^x}P_{2k}}{k!k!} \times 2^{x-2k} = \dfrac {(2x)!}{x!x!}$, where $n$ is the largest integer lower than or equal to $\dfrac{x}{2}$.
If it is true, please share some references or hint on how I should proceed with the proof.
You can define a bijection between the two. Take a string of length $2x$ and break it into pairs of letters. Replace $AA$ with $A$, $BB$ with $B$, $AB$ with $C$ and $BA$ with $D$. You create a matching string of length $x$ with an equal number of $A$s and $B$s.