Counting partitions of an $n$-element set.

Question

I want to count the number of ways I can partition a set such that $n-1$ blocks have a single element and $1$ block has $2$ elements. I was thinking that $n\choose 2$ would do it. Any thoughts?

Answer 1

Yup, you're right. You pick the block you want to have $2$ elements in, and the others follow automatically. If it's of any use to you, note that you can write

$${n\choose 2}=\tfrac12n^2-\tfrac12n$$