Let $a(n, k)$ be the number of permutations of length n with k cycles in which the entries 1 and 2 are in the same cycle. Let $t(n, k) = c(n, k) − a(n, k)$ be the number of permutations of length n with k cycles in which the entries 1 and 2 are not in the same cycle. Prove that $a(n, k) = t(n, k + 1)$ for all $k ≤ n − 1.$
EDIT: I've tried directly summing the generating functions for (,),(,), and we got $x^2(+1)...(+−1)$, but we don't know where to go from there. We also tried using the fact that $t(n, k) = c(n, k) − a(n, k)$ to perhaps get some sort of relation that makes more sense, but the only relations that we can make rely on using the fact we're trying to prove, which is sort of counterproductive.
Hint: Let $\sigma$ be one of the permutations counted by $a(n,k),$ then $$\sigma = (1\, a_1\, \cdots\, a_i\, 2\, b_1\, \cdots \, b_j)c_2\cdots c_k,$$ where $c_i$ are cycles and $\{a_m,b_m\}$ are numbers. What if you define a function $$\varphi (\sigma )=(1\, a_1\, \cdots\, a_i)(2\, b_1\, \cdots \, b_j)c_2\cdots c_k.$$ Is this a bijection?