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- Path: sparky!uunet!ogicse!das-news.harvard.edu!husc-news.harvard.edu!blom
- From: blom@husc15.harvard.edu
- Newsgroups: sci.math
- Subject: The Mathematics of Secret Santa
- Message-ID: <1992Dec17.220416.18649@husc15.harvard.edu>
- Date: 18 Dec 92 03:04:16 GMT
- Article-I.D.: husc15.1992Dec17.220416.18649
- Organization: Harvard University Science Center
- Lines: 36
-
- What is the probability that in a Secret Santa arrangement of n people, there
- is at least one pair of participants with each other as Secret Santas?
-
- No participant may have himself.
-
- The total number of Secret Santa arrangements turns out to be the subfactorial
- of the number of participants. Here, the subfactorial of the number n is
- defined as
- k
- n (-1)
- !n = n! sum ------
- k=0 k!
-
- For example, !5 = 5! (1/0! - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!)
- = 120 (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
- = 44.
-
- The subfactorial, I think, should be the denominator of the fraction for the
- probability with n people, but I can't figure out the numerator.
-
- ??
- P(n) = ----
- !n
-
- For the record, here are the first few values of P(n)*!n, which should be the
- numerator of the fraction:
-
- !2*P(2) = 1
- !3*P(3) = 0
- !4*P(4) = 3
- !5*P(5) = 20
- !6*P(6) = 105
- !7*P(7) = 504
-
- Eric Blom
- Harvard U
-