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Groups and Group Actions(2014) page 22-23:
Example 68 (i) How many 5-cycles are there in $S_{11}$ ?
(ii) How many permutations in $S_{8}$ have a cycle decomposition type of two 3-cycles and one 2-cycle?
Solution. (i) 5-cycles in $S_{11}$ are of the form $(a b c d e)$. There are 11 choices for what $a$ might be, 10 for $b, 9$ for $c, 8$ for $d$ and 7 for $e$. However we need to remember that
$$
(a b c d e)=(b c d e a)=(c d e a b)=(d e a b c)=(e a b c d)
$$
and so there answer is
$$
\frac{11 \times 10 \times 9 \times 8 \times 7}{5}=11088 .
$$
(ii) Permutations in $S_{8}$ that decompose to two 3-cycles and a 2-cycles are of the form
$$
(a b c)(d e f)(g h) .
$$
There are $8 !$ ways of filling in these brackets as $a \ldots h$ but we need to remember that the above permutation also equals
$$
(b c a)(d e f)(g h)=(a b c)(e f d)(g h)=(a b c)(d e f)(h g)=(d e f)(a b c)(g h) .
$$
The first three are equivalent rewritings that come from cycling elements within cycles whilst the last one comes from permuting two equal length cycles. Hence the number of such permutations is
$$
\frac{8 !}{3 \times 3 \times 2 ! \times 2}=\frac{40320}{36}=1120 \text {. }
$$
(i') Note that we could have tackled the 5-cycle question in this manner as well. Thinking of 5-cycles in $S_{11}$ as having cycle type
$$
(a b c d e)(f)(g)(h)(i)(j)(k)
$$
we see that there are $11 !$ ways of filling these brackets as $a \ldots k$ but $5$ ways of cycling $a \ldots e$ and $6 !$ ways of permuting the equal length cycles $(f) \ldots(k)$ Hence the answer is
$$
\frac{11 !}{5 \times 6 !}=11088 \text {. }
$$
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Proposition 69 In $S_n$ there are$$\frac{n !}{\left(l_{1}^{k_{1}} \times l_{2}^{k_{2}} \times \cdots \times l_{r}^{k_{r}}\right)\left(k_{1} ! \times k_{2} ! \times \cdots \times k_{r} !\right)}$$permutations with a cycle type of $k_1$ cycles of length $l_1$, $k_2$ cycles of length $l_2$,⋯, $k_r$ cycles of length $l_r$. This decomposition includes 1-cycles so that$$k_{1} l_{1}+k_{2} l_{2}+\cdots+k_{r} l_{r}=n$$
Example 70 How many permutations of each cycle type are there in $S_{7}$ ?
Solution. The table below contains the various numbers. We need to consider the various ways in which 7 can be composed as other integers.
type |
working |
# |
type |
working |
# |
type |
working |
# |
7 |
$7 ! / 7$ |
720 |
$4+2$ |
$7 ! /(4 ×2)$ |
630 |
3 |
$7 ×6 ×5\over3$ |
70 |
6 |
$7 ! / 6$ |
840 |
4 |
$7 ×6 ×5 ×4\over4$ |
210 |
$3 ×2$ |
$7 ×6 ×5 ×4 ×3 ×2\over2 ×2 ×2 ×3 !$ |
105 |
$5+2$ |
$7 ! /(5 ×2)$ |
504 |
$2 ×3$ |
$7 ×6 ×5 ×4 ×3 ×2\over3 ×3 ×2 !$ |
280 |
$2 ×2$ |
$7 ×6 ×5 ×4\over2 ×2 ×2 !$ |
105 |
5 |
$7 ×6 ×5 ×4 ×3\over5$ |
504 |
$3+2 ×2$ |
$7 !\over3 ×2 ×2 ×2 !$ |
210 |
2 |
$7 ×6\over2$ |
21 |
$4+3$ |
$7 ! /(4 ×3)$ |
420 |
$3+2$ |
$7 ×6 ×5 ×4 ×3\over3 ×2$ |
420 |
$e$ |
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1 |
|
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