# Determine how many distinct pairs

Given a set of distinct elements $\{x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8 \}$ how many distinct (order doesn't matter) pairs $(y,z)$ is it possible to obtain?

I just got stuck here. I believe it is $\binom{8}{2}$ but in the book it says $36$.

• just as a side note: this is the difference between combin and combina in excel. You did combin (combination) and the result should be used with combina (where repetition allows) 28+8=36 – adhg Jun 7 '18 at 22:42

Hint. The number $\binom{8}{2}$ enumerates the pairs $(y,z)$ where $y<z$, but the pairs can also be of the form $(y,z)$ with $y=z$: $$(x_1,x_1),(x_2,x_2),\dots,(x_8,x_8).$$