3
$\begingroup$

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 $36$.

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$ – adhg Jun 7 '18 at 22:42
3
$\begingroup$

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 $y=z$: $$(x_1,x_1),(x_2,x_2),\dots,(x_8,x_8).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.