Difference between revisions of "2016 AMC 10A Problems/Problem 1"
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<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math> | <math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math> | ||
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| + | ===Solution=== | ||
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| + | <math>\frac{11!-10!}{9!} = \frac{9!(10 \cdot 11 -10)}{9!} = 10 \cdot 11 - 10 =\boxed{\textbf{(B)} 100}</math> | ||
Revision as of 17:51, 3 February 2016
What is the value of 
?
Solution
