, 2011

Let $D$ be the set of 25-digit numbers with the same number of 1s and 2s, and let $R$ be the set of 50-digit numbers with 25 digits 1 and 25 digits 2. We define a bijection between the two sets.

Let $d = d_1d_2\dots d_{25} \in D$. We define a function on the digits of $d$ as follows:
$$f(d_i) = \begin{cases} 11 & \text{if } d_i = 1 \\ 22 & \text{if } d_i = 2 \\ 12 & \text{if } d_i = 3 \\ 21 & \text{if } d_i = 4 \end{cases}$$
By replacing each digit $d_i$ in $d$ with $f(d_i)$ we obtain a number in $R$; indeed, if $d_i$ equals 3 or 4, then $f(d_i)$ contains the same number of 1s and 2s, and since there are equal numbers of 1s and 2s in $d$, there will be equal numbers of the digit pairs 11 and 22. Hence each number in $d$ corresponds to a number in $R$ - clearly two different numbers in $d$ will correspond to two different numbers in $R$.

Conversely, using the inverse of $f$, we can map each number in $R$ to a number in $D$. A similar argument shows that this mapping is well-defined and injective as well, showing that the two sets have an equal number of elements.