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Revision as of 15:29, 9 February 2019

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 1

Factoring out $10!$ from the numerator and cancelling out $9!$ from the numerator and the denominator, we have \[\frac{11!-10!}{9!} = \frac{11 \cdot 10! - 1 \cdot 10!}{9!} =  \frac{(10!) \cdot (11 - 1)}{9!} = 10 \cdot 10 =\boxed{\textbf{(B)}\;100}.\]

Solution 2

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


Solution 3

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

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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Problem 2
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All AMC 10 Problems and Solutions
2016 AMC 12A (ProblemsAnswer KeyResources)
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Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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