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Topic: Homework question: permutations
Replies: 12   Last Post: Oct 21, 2001 8:04 PM

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Robin Chapman

Posts: 2,247
Registered: 12/6/04
Re: Homework question: permutations
Posted: Oct 15, 2001 4:17 AM
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"Fred Galvin" <> wrote in message

> S_n is the group of all permutations of {1,2,...,n}.
> If n is an even number, the following probabilities are equal:
> (a) the probability that a random element of S_n has odd order;
> (b) the probability that a random element of S_{n+1} has odd order;
> (c) the probability of getting equal numbers of heads and tails in n
> independent tosses of a fair coin.
> This came up in a homework problem in my combinatorics class. It's
> easy enough to verify the equalities by mindless computation with
> generating functions, but I think they must have a simple explanation.
> What is it?

For (a) and (b) let o_n denote the number of odd order permutations
in S_n. Then it's easy to prove
o_{n+1} = o_n + n (n-1) o_{n-1}.
Note that o_n counts the permutations of odd order in S_{n+1}
fixing n+1 while (n-1)o_{n-1} counts those taking n+1 to j (1 <= j <=n)
[delete n+1 and j from a permutation, and note that these could have
occurred in n-1 places].
Now it's easy to induct that o_{n+1} = n o_n for even n and
o_{n+1} = (n-1)o_n for odd n.

Dunno about (c) :-(

Robin Chapman

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