Chapter 12 - Counting and Probability - Section 12.2 Permutations and Combinations - 12.2 Assess Your Understanding - Page 876: 61

$15$

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. Hence here $_{6}C_2=\frac{6!}{(6-2)!2!}=15$ There are $15$ different teams possible.