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I would like to know why do we need to divide by 2! while considering pairs in p.11 of Enigma Machine.The article says that:

"Number of ways to choose four letters from 26 letters is $\frac {{26 \choose 4}{4 \choose 2}{2 \choose 2}}{2!}$"

but I can't get why do we need to divide by 2! since I think we're all have already done all the divisions to consider the combinations of the letters in the Enigma Machine.

justin
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2 Answers2

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$\binom{24}{4}\binom 42\binom 22/2!$ counts selections of 4 from 24 letters (without replacement) into two pairs when we consider permutation of the pairs to be indistinct.

That is: we could select, for instance, ABCD, then divide them into pairs in three ways; which is $\binom 42/2!$ because, those pairings on the left column of the table are indistinct from those on the right of the same row. $$\begin{array}{c:c}AB,CD & CD, AB\\ \hdashline AC,BD & BD,AC\\ \hdashline AD, BC& BC, AD\\ \end{array}$$

Graham Kemp
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There are six ways to choose two out of four: $$ \begin{array}{cc} \text{chosen} & \text{unchosen} \\ AB & CD \\ AC & BD \\ AD & BC \\ BC & AD \\ BD & AC \\ CD & AB \end{array} $$ But if we're just interested in the number of ways to partition a set of four things into two subsets of two each, then choosing $AB$ and leaving $CD$ unchosen is the same as choosing $CD$ and leaving $AB$ unchosen. Likewise $AC$ versus $BD$ becomes the same thing as $BD$ versus $AC$, and $AD$ versus $BC$ becomes the same as $BC$ versus $AD$, so there are only half as many: $$ \begin{array}{c} AB/CD \\ AC/BD \\ AD/BC \end{array} $$

Michael Hardy
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