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Let $S_4$ be the set of all permutations on $4$ symbols and $A$ a subset of $S_4$ such that $$A = \{f\in S_4 : f \text{ is a 3-cycle}\},$$ then $|A|= \mathord{?}$

(included from comments)
I think number of $3$-cycles of $S_4$ is ${}^4C_3 \times 2 =8$ so order of $A$ is $8$

...since $4 <3+3$ there are $8\ \ 3$-cycles because there are ${}^4C_3$ ways to select the $3$ elements and $2$ times to orient each cycle

...is it correct?

Joffan
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Halima.Khatun
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  • i think number of 3 cycle of S4 is 4C3 X 2 =8 so order of A is 8 ..is it correct – Halima.Khatun Dec 30 '16 at 05:22
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    Yes, it is correct. Are you able to explain why it is correct? Why $\binom{4}{3}$? Why $2$? Why multiply them? – JMoravitz Dec 30 '16 at 05:26
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    You should include the information that you commented in the body of your question. People on this site like to see that you have attempted the problem yourself. – pseudoeuclidean Dec 30 '16 at 05:26
  • It should be $P(4,3)/3=8$, since each $3$ cycle has three different but equivalent forms (e.g., $(123)=(231)=(312)$). – Zuriel Dec 30 '16 at 05:26
  • It is correct. But could you please explain why you thought that (maybe by editing your question)? Sometimes even just explaining is helpful to verifying your thinking! – Chas Brown Dec 30 '16 at 05:27
  • @zuriel notice that $P(4,3)/3=\binom{4}{3}\cdot 2 = 8$ is just a different way of writing the same answer – JMoravitz Dec 30 '16 at 05:27
  • @JMoravitz， thanks for pointing out. I did not see the OP's reasoning but the answer is correct. – Zuriel Dec 30 '16 at 05:28
  • since 4 <3+3 there are 8 3-cycles because there are 4C3 ways to select the 3 elements and 2 times to orient the cycle. – Halima.Khatun Dec 30 '16 at 05:34
  • Added your thoughts into the question. – Joffan Dec 30 '16 at 06:14

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