Difference between revisions of "2016 AMC 12A Problems/Problem 1"

Latest revision as of 11:44, 4 February 2016

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-==Problem==
+#REDIRECT [[2016 AMC 10A Problems/Problem 1]]
-What is the value of <math>\frac{11!-10!}{9!}</math>?
-<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math>
-==Solution==
-<cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!}</cmath>
-<cmath>=\frac{11\cdot 10\cdot 9!}{9!} - \frac{10\cdot 9!}{9!}</cmath>
-<cmath>=(11\cdot 10) - 10</cmath>
-<cmath>\boxed{\textbf{(B) } \, 100}</cmath>
-==See Also==
-{{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}}
-{{MAA Notice}}