Let A = {1, 2,..., 10}. How many three-element subsets of A contain at least two consecutive integers?
I believe there are $\displaystyle \tbinom{10}{3}$ total 3-subsets of A. To find the subsets containing at least two consecutive integers, I thought to subtract from the total all subsets that do not contain consecutive integers.
I had some trouble understanding the general formula for determining the number of size-k subsets of a size-n set that don't contain consecutive integers, but this explanation helped.
Anyway, that gives me $\displaystyle \tbinom{10}{3}- \tbinom{n-k+1}{k}= 120 - \tbinom{10-3+1}{3}= 64$.
Did I miss anything?
Thanks!