Posted by: sandeepchrao | January 2, 2009

## Applications of 152 in Secret Santa

(assuming you assign ever person a number)

(9  4  6  7  13)(2  12  3  5   11  1  10  8)         A permutation with 1 cycle of 5 and 1 cycle of 8

The previous year, we had a Secret Santa with the same group and the gift cycle structure had the same structure as this permutation

( 1  7  13 4  2  11  9  12 3  8  5  10  6)

One full cycle of 13 is rare among all the cycle structures possible with the 13! possible permutations. I think we were pretty lucky.

Anyway, thought this was interesting. Happy Holidays.

## Responses

1. I was reading your blogs with interest and came across this Secret Santa permutation question. At first I thought that, yes a full cycle was rare. Then I started quantifying it.
For n people, there are n! permuations of slips of names they can get (allowing for the sad scenario of people getting presents for themselves).
How many of these are complete cycles? WLOG assume the first number in the cyclic permutation is 1. Then we are left with n-1 remaining numbers in any order. So there are (n-1)! possible full-cycle permutations. Hence 1/n of the possible secret santa permutations is a full cycle. More if self-presenting is ruled out! I was surprised at how often these full cycles arose. I even hand worked a few examples to convince myslef I hadn’t made a silly mistake.
Thanks for a thought provoking few minutes! Now to ponder the number of ways of getting various cycle-combinations…

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