TITLE: Finding the cardinality of Ai and summation
QUESTION [1 upvotes]: Let A = {1, 2, 3,..., n} (i.e. a set with n elements). Recall that a permutation z is an invertible function. Let Sn denote the set of all permutations on A.
Let Ai = {z ∈ Sn | z(i) = i},
where
i = 1,...,n.
That is, Ai consists of all the permutations that fix the i-th element of A.
(a) Find |Ai|
(b) Find  $$S_1 = \sum_{i=1}^n |A_i|$$

For (a) I got the cardinality to be n! because the i-th element is fixed but I'm not too sure about this.
So for (b) would I be doing the summation of n!?

REPLY [1 votes]: Close, but $|A_i| = (n-1)!$ since the $i -th$ element is fixed (since we can think of this as the permutation group acting on $(n-1)$ elements). 
Then $\sum_{i=1}^n |A_i| = \sum_{i=1}^n (n-1)! = n\cdot ((n-1)!) = n!$
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