# Calculate all shuffle of permutation $S_4$ [closed]

How can I Calculate all shuffle of permutation $S_4$? Thank you

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What do you mean by shuffle here? –  Berci Feb 18 '13 at 18:13
And what does this have to do with manifolds or exterior algebras? (I would tag it as, say, group-theory, abstract-algebra, and permutations.) –  anon Feb 18 '13 at 18:14
Pick a piece of paper (or text editor) and write down all the $24$ permutations, and collect the shuffle ones whatever it means. Most probably, meanwhile you will get also the answer on how to calculate these. –  Berci Feb 18 '13 at 18:30
@AlexanderGruber, you both voted to close as not reasonably answerable and answered this question. Did you change your mind? I am wondering if this really should have been closed (and downvoted so much) to begin with, and if we could benefit from reopening it. –  anon Mar 3 '13 at 23:07
What you mean? @anon –  B11b Mar 3 '13 at 23:09

## closed as not a real question by Hagen von Eitzen, Thomas, Alexander Gruber♦, Henry T. Horton, NamelessFeb 18 '13 at 19:37

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A shuffle of a cycle $a=(a_1 a_2 \ldots a_n)$ with a disjoint cycle $b=(b_1 b_2 \ldots b_m)$ is some cycle $(c_1 c_2 \ldots c_{m+n})$ where each $c_i$ is equal to an $a_i$ or $b_i$ and the $a$'s and $b$'s are kept in their original order. For example, some possible shuffles of $(\color{red}{123})$ with $(\color{blue}{456})$ are: $$(\color{red}1\color{blue}4\color{red}2\color{blue}5\color{red}3\color{blue}6), (\color{red}{12}\color{blue}{45}\color{red}3\color{blue}6) , (\color{red}{123}\color{blue}{456}),(\color{blue}4\color{red}1\color{blue}5\color{red}2\color{blue}6\color{red}3), (\color{blue}{45}\color{red}{12}\color{blue}6\color{red}3)$$ The set of shuffles of $a$ and $b$ is denoted $\operatorname{sh}(a,b)$ and has size ${n+m \choose n}$.
So let's calculate the shuffles in $S_4$, which I'll write $\operatorname{sh}(S_4)$. We can't shuffle the identity with anything, and if we pick a $3$ or $4$ cycle we can't pick a disjoint cycle to shuffle it with. So, let's shuffle two $2$-cycles: $(\color{blue}{b_1b_2}\color{red}{a_1a_2})$ $(\color{blue}{b_1}\color{red}{a_1}\color{blue}{b_2}\color{red}{a_2})$ $(\color{blue}{b_1}\color{red}{a_1a_2}\color{blue}{b_2})$ $(\color{red}{a_1}\color{blue}{b_1}\color{red}{a_2}\color{blue}{b_2})$ $(\color{red}{a_1a_2}\color{blue}{b_1b_2})$
Since we can pick $a$ and $b$ to be whatever we want, this is just the set of $4$-cycles. The identity shuffled with something is defined to be the identity, so we have that $\operatorname{sh}(S_4)$ is the set of $4$-cycles in $S_4$.