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2016 AMC 10A Problems/Problem 1

Revision as of 17:58, 3 February 2016 by Benq (talk | contribs)

Problem

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

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

Solution

\[\frac{11!-10!}{9!} = \frac{9!(10 \cdot 11 -10)}{9!} = 10 \cdot 11 - 10 =\boxed{\textbf{(B)} 100}\]

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