Difference between revisions of "2016 AMC 10A Problems/Problem 1"

Revision as of 18:28, 26 January 2020

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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All AMC 10 Problems and Solutions

2016 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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 ==Solution 1==
-Factoring out <math>10!</math> from the numerator and cancelling out <math>9!</math> from the numerator and the denominator, we have <cmath>\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} =  \frac{(10!) \cdot (11 - 1)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100}.</cmath>
+<math>\frac{11!-10!}{9!}=\frac{11\cdot10!-10!}{9!}=\frac{100\cdot9!}{9!}=100</math>
+<math>B</math>
 ==Solution 2==