How many elements of order 2 does the symmetric group S_4 contain

These come in two varieties:

(i) (a b), for some distinct a, b in {1, 2, 3, 4}.
There are 4 choices for a and 3 choices for b
Remembering that (a b) = (b a), there are (4 * 3)/2 = 6 such elements.

(ii) (a b)(c d), with a,b,c,d distinct elements from {1, 2, 3, 4}.

There are (4 * 3)/2 elements for a, b; this fixes c, d (via (2 * 1)/2 = 1).
We have to divide by 2, because (ab)(cd) = (cd)(ab)
==> (4 * 3)/(2 * 2) = 3 such elements.

By (i) and (ii), there are 6 + 3 = 9 elements of order 2 in S_4.

I hope this helps!