TITLE: How many 3-subsets of $\{1,2,\ldots,10\}$ contain at least two consecutive integers?
QUESTION [6 upvotes]: 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!

REPLY [2 votes]: You could count the number of $3$ element subsets that have no consecutive integers.
The first number $n_1$ is taken from $1,...,6$ (any higher and there must be a consecutive pair). The next number $n_2$ is taken from $n_1+2,...,8$, and the last number $n_3$ is taken from $n_2+2,...,10$.
Summing gives
\begin{eqnarray}
\sum_{n_1=1}^6 \sum_{n_2=n_1+2}^8 \sum_{n_3=n_2+2}^{10} 1 &=& \sum_{n_1=1}^6 \sum_{n_2=n_1+2}^8 (9-n_2) \\
&=& \sum_{n_1=1}^6 \sum_{n_2=n_1}^6 (7-n_2) \\
&=& \sum_{n_1=1}^6 \sum_{n_2=1}^{7-n_1} (8-n_2-n_1) \\
&=& \sum_{n_1=1}^6 ((8-n_1)(7-n_1)-\frac{1}{2}(7-n_1)(8-n_1)) \\
&=& \sum_{n=1}^6 \frac{1}{2}(8-n)(7-n) \\
&=& \frac{1}{2} \sum_{n=1}^6 (n^2-15n+56) \\
&=& \frac{1}{2} (91-15\frac{1}{2}(6)(7)+6(56))\\
&=& 56
\end{eqnarray}
Since $\binom{10}{3} = 120$, the answer is $120-56= 64$.