2016 AMC 10A Problems/Problem 1

Problem

What is the value of $\dfrac{11!-10!}{9!}$ ?

$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

$\frac{11!-10!}{9!}=\frac{11\cdot10!-10!}{9!}=\frac{100\cdot9!}{9!}=100$ $B$

We can use subtraction of fractions to get $\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!} = 110 -10 = \boxed{\textbf{(B)}\;100}.$

Factoring out $9!$ gives $\frac{11!-10!}{9!} = \frac{9!(11 \cdot 10 - 10)}{9!} = 110-10=\boxed{\textbf{(B)}~100}$ .

2016 AMC 10A (Problems • Answer Key • Resources)
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2016 AMC 12A (Problems • Answer Key • Resources)
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