Talk:Partial permutation

Latest comment: 10 years ago by Arthur Rubin in topic Recurrence relation

Recurrence relation

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I don't see it. The argument is fairly complex, and doesn't seem correct.

I get

 

Counting as follows. Let the n-element sets be A and B, with distinguished elements aA and b in B.

P(n−1) where neither a or b is active. (Sets being A−{a} and B−{b}.)
P(n−1) where a maps to b. (Sets being A−{a} and B−{b})
(n−1) P(n−1) where a maps somewhere but not to b (Sets being A−{a} and B−{f(a)}.)
(n−1) P(n−1) where b maps somewhere but not to a. (Sets being A−{f−1(b)} and B−{b}.)
less permutations where a and b map somewhere, but not to each other, which are counted in both of the last two sets.
(n−1)2 P(n−2) (Sets being A−{a,f−1(b)} and B−{b,f(a).}

But I think this is complicated enough to need a reference. — Arthur Rubin (talk) 03:53, 15 August 2014 (UTC)Reply