Answer
$15$
Work Step by Step
If we want to choose $k$ elements out of $n$ disregarding the order, not allowing repetition, we can do this in $_{n}C_k=\frac{n!}{(n-k)!k!}$ ways.
The order doesn't matter here when choosing the jokes, thus we have to use combinations. We choose $2$ out of $6$. Thus $_{6}C_{3}=\frac{6!}{(6-2)!2!}=15$