Maths Olympiad Prep

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HMMT February

Combinatorics Difficulty 5.1 AIME, harder Find the answer

A permutation of a finite set is a one-to-one function from the set to itself; for instance, one permutation of {1,2,3,4}\{1,2,3,4\} is the function π\pi defined such that π(1)=1,π(2)=3\pi(1)=1, \pi(2)=3, π(3)=4\pi(3)=4, and π(4)=2\pi(4)=2. How many permutations π\pi of the set {1,2,…,10}\{1,2, \ldots, 10\} have the property that π(i)≠i\pi(i) \neq i for each i=1,2,…,10i=1,2, \ldots, 10, but π(π(i))=i\pi(\pi(i))=i for each ii?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

For each such π\pi, the elements of {1,2,…,10}\{1,2, \ldots, 10\} can be arranged into pairs {i,j}\{i, j\} such that π(i)=j;π(j)=i\pi(i)=j ; \pi(j)=i. Choosing a permutation π\pi is thus tantamount to choosing a partition of {1,2,…,10}\{1,2, \ldots, 10\} into five disjoint pairs. There are 9 ways to pair off the number 1, then 7 ways to pair off the smallest number not yet paired, and so forth, so we have 9⋅7⋅5⋅3⋅1=9459 \cdot 7 \cdot 5 \cdot 3 \cdot 1=945 partitions into pairs.

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