Number of permutations with cycle shape (2,2) in S5

Question

I probably need permutations of form $(ab)(cd)$. If I had a 3 cycle $(abc)$ I would use something we call arrangements. For example arrangements of $n$ choose $k$ is just $\frac{n!}{k!}$. In this case, I would get $5\times4 = 20$ permutations with cycle shape $(3)$. Now because I have both $(ab)$ and $(cd)$ I am not sure how to proceed. I thought I take the case where b=c and have 20 permutations as before and I subtract 5! divided by 4! which is 5 (when a,b,c,d are distinct) and get $20-5=15$. Does that have any logic? Thanks!

Answer 1

I am not sure I follow your reasoning, but here is a different approach. Note (in case you are unfamiliar with the notion) that I am using $\binom nr =\frac {n!}{r!(n-r)!}$ which is the number of ways of choosing $r$ objects from $n$ if the order does not matter.

I can choose the first pair in $\binom 52$ ways and the second pair from the remaining three numbers in $\binom 32$ ways. But the order of the pairs doesn't matter - I get the same group element - so I must also divide by $2!=2$ because there are two pairs.

I get $\frac 1{2!} \binom 52 \binom 32=\frac 12\times 10 \times 3 =15$.

That at least gives the same number, which may be coincidence.

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Just to illustrate how this point of view can help, consider elements of the form $(2,2,2)$ in $S_7$. Here there are three $2-$cycles, and because the order of these does not matter, I must divide by $3!$

The number is $\frac 1{3!}\binom 72\binom 52 \binom 32$ which I can write down straight away and then calculate.

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To go a little further:

If you write this down in factorials you will see a fair amount of cancellation. In fact this leads to the expression of the appropriate multinomial coefficient from the product of binomials.

In the general case there will be factors like $\frac 1{3!}$ to eliminate any double-counting of transpositions, $3-$cycles etc

Answer 2

It sounds as though you are trying to count the number of permutations in $S_5$ whose disjoint cyclic decomposition has two $2$-cycles.

Approach with multiplication principle.

• Pick which four elements appear in your two $2$-cycles
• There will be a unique "smallest" picked element. Pick which of the other picked elements appears in its cycle. The remaining two form their own cycle.

This can be done in $\binom{5}{4}\cdot 3 = 15$ ways.

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As an aside, for counting the number of permutations of $S_5$ whose cyclic decomposition results in exactly one $3$-cycle and the rest of the elements being fixed points, this can be done similarly by picking which three elements appear in the cycle, finding which is the smallest, and then placing the remaining elements around the cycle, giving $\binom{5}{3}\times 2 = 20$.

That you got the answer of $20$ appears to be a coincidence. If we were talking about $S_6$ the correct approach would have given $\binom{6}{3}\times 2 = 40$ while your attempt would have gotten an answer of $6\times 5\times 4 = 120$ Similarly, your attempt for calculating the permutations with two disjoint $2$-cycles looks incorrect as well despite coincidentally arriving at the correct final answer.

Answer 3

For an element of the form $(ab)(cd)$ in $S_5$, there are $5$ possible choices for $a$ and $4$ remaining possible choices for $b$, which would be $20$ possible choices for $(ab)$, but since $(ab)=(ba)$, divide by $2$ to get $10$. After $a$ and $b$ are chosen, there are $3$ remaining possible choices for $c$ and $2$ remaining possible choices for $d$, which would be $6$ possible choices for $(cd)$, but again divide by $2$ to get $3$. Thus, there are $10\times3=30$ possible choices for $(ab)(cd)$, but divide by $2$ because $(ab)(cd)=(cd)(ab)$, so the answer is $15$.

Answer 4

$\dfrac {P^5_2}2\cdot\dfrac {P^3_2}2=10\cdot3=30$. Then divide by $2$ because the two $2$-cycles can occur in two different orders.

So $15$.