Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1050: 78

$900$

We know that $n!=1 \cdot 2 \cdot 3 .....(n-1)n$ Thus, we have $\dfrac{900!}{899!}=\dfrac{1 \cdot 2 \cdot 3 .....899 \cdot 900}{899!}$ $=\dfrac{(1 \cdot 2 \cdot 3 .....899) \cdot 900}{899!}$ $=\dfrac{899! \cdot 900}{899!}$ $=900$