Chapter 11 - Review Exercises - Page 871: 55

$4005$

Binomial coefficient can be found as: $\dbinom{n}{r}=\dfrac{n!}{r!(n-r)!}$ Thus, $\dbinom{90}{2}=\dfrac{90!}{2!(90-2)!}$ or, $=\dfrac{ 90 \cdot 89 \cdot 88!}{2!(88)!}$ or, $=\dfrac{ 90 \cdot 89 \cdot 88!}{2 \cdot 1(88)!}$ or, $=4005$