# Chapter 11 - Counting Methods and Probability Theory - 11.2 Permutations - Exercise Set 11.2: 59

900

#### Work Step by Step

$n!=n(n-1)(n-2)\cdots(3)(2)(1)$ and, by definition, $0!=1$ ----------------- Most calculators can display up to $69!\approx 10^{98}$. Anything over results in an error. 900! and 899! are huge numbers. We employ a different approach: $900!=900\times 899\times 898\times...\times 2\times 1$ $=900\times 899!$ So, we write $\displaystyle \frac{900!}{899!}=\frac{900\times 899!}{899!}$ ... reduce the fraction by $899!$... $=900$

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