0

Any permutation is a product of cycles. For example, the permutation 351642 $(3 \Rightarrow 1, 5 \Rightarrow 2, 1 \Rightarrow 3, 6 \Rightarrow 4, 4 \Rightarrow 5, 2 \Rightarrow 6 )$ can be written as $(31)(2645)$ How many permutations of $1,...,8$ are a product of a 1-cycle, two 2-cycles, and a 3-cycle?

I know the answer is $\frac{8!}{2^{3}3!} × 2$

Using the formula $\frac{(mn)!}{(n!)^mm!}$ where a set of $mn$ objects can partitioned into $m$ set of $n$ size. I know the numerator comes from the amount of permutations and the $3!$ in the denominator comes from the amount of sets we want ($m$). What I don't understand is why the $2$ is to the power of $3$ and why the entire thing is multiplied by $2$.

Casteels
  • 11,292
  • 4
  • 27
  • 38
jeremysanchez50
  • 303

1 Answers1

1

There are $\frac{8!}{2^2\times 3!}$ ways to choose two pairs and one triple out of $8$ numbers. These tuples determine the cycles. The remaining number is the $1$-cycle.

The result is right, but I have no idea why it is written in the given way.

Peter
  • 84,454
  • Why do they multiply the entire thing by 2 though? – jeremysanchez50 Nov 07 '15 at 19:31
  • And what is the meaning of the $2^3$ ? – Peter Nov 07 '15 at 19:34
  • Perhaps I'm mistaken, but I thought that for a permutation of n elements of cycle type ${1^{c_1}, 2^{c_2}...}$, the formula was $n!/((c_1)!(c_2)!...1^{c_1}*2^{c_2}...)$. To my understanding, the answer shouldn't have that factor of 2. EDIT: Wait, nevermind, I missed that that was 3!, not 3, that is correct. Strange way of writing it, however. – Kevin Long Nov 07 '15 at 20:22