Question

Combinations- how to show that 10C2=10C8

Answer 1

By definition,

${}^nC_k=\dbinom nk=\dfrac{n!}{k!(n-k)!}$

We have

$\dbinom{10}2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10!}{(10-8)!8!}=\dbinom{10}8$

and so

${}^{10}C_8={}^{10}C_2$