Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.5 The Binomial Theorem - 11.5 Assess Your Understanding - Page 857: 10

$4950 $

According to the binomial theorem, we have: $\displaystyle{n\choose k}=\dfrac{n!}{(n-k)! \ k!}$. Substitute $100$ for $n$ and $98$ for $k$ in the above formula. Therefore, $\dbinom{100}{98}=\dfrac{100!}{98! \ (100-98)!} \\ =\dfrac{100!}{2! \ 98!}\\=\dfrac{100 \cdot 99 \cdot 98!}{98 ! (2 \cdot 1) }\\= 4950 $