Number of permutations with $n-2$ fixed-points and one $2$-cycle same as number of permutations with $n-6$ fixed-points and three $2$-cycles. I have a question to an exercises which i cannot solve:
Determine all $n\geq 6$ for which the following statement is correct: There are as many permutations with $n-2$ fixed-points and one $2$-cycle as permutations with $n-6$ fixed-points and three $2$-cycles.
I'm trying to find a bijection between the two sets of those permutation but i can't find a function.
Any help is highly appreciated. Thanks!
Edit: Made a mistake in the problem of the exercise. The bold text is edited.
 A: The proposition is true only when $n=6.$ (The question has been edited so that $n=1$ or $n=0$ need no longer be mentioned.)
The number of ways to choose a $2$-cycle is $\dbinom n 2.$
The number of ways to choose three $2$-cycles is $\dbinom n 6 \cdot 15,$ since there are $15$ ways to make a set of $6$ into three $2$-cycles.
So you need $\dbinom n 6 \cdot 15 = \dbinom n 2.$
$$
\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)} {6\cdot5\cdot4\cdot3\cdot2\cdot1} \cdot 15 = \frac{n(n-1)} 2.
$$
$$
(n-2)(n-3)(n-4)(n-5) = 24. \tag 1
$$
If $n>6$ then the left side of $(1)$ is $>24.$ If $2\le n\le 5$ then the left side of $(1)$ is $0.$
A: The number of permutations of $\{1,\dots,n\}$ with $c_k$ cycles  of size $k$ for $k=1,\dots, n$, is given by
$$\frac{n!}{(1^{c_1}\cdot 2^{c_2}\cdots n^{c_n})(c_1!\cdot c_2!\cdots c_n!)}.$$
(see Number of permutations for a cycle-type). Speaking briefly, we have $n$ positions: the first $c_1$ are the $1$-cycles (fixed points), the next $2c_2$ are the $2$-cycles an so on. We arrange the $n$ elements in $n!$ ways, and for each $k$ we divide by $c_k!$ because the order of the $k$-cycles is indifferent, and we divide also by $k^{c_k}$, because a shift in a $k$-cycles is indifferent.
Now we find in your example the cardinalities of the two sets of permutations. 
1) $(n-2)$-cycles and one $2$-cycle:
$$\frac{n!}{(1^{n-2}\cdot 2^{1})((n-2)!\cdot 1!)}=\frac{n(n-1)}{2}$$
2) $(n-6)$-cycles and three $2$-cycle:
$$\frac{n!}{(1^{n-6}\cdot 2^{3})((n-6)!\cdot 3!)}=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{48}$$
When are they equal?
