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

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==Solution==
 
==Solution==
  
<cmath>\frac{11!-10!}{9!} = \frac{9!(10 \cdot 11 -10)}{9!} = 10 \cdot 11 - 10 =\boxed{\textbf{(B)}\;100.}</cmath>
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Factoring out the <math>10!</math> from the numerator and cancelling out the <math>9!</math>s in the numerator and denominator, we have: <cmath>\frac{11!-10!}{9!} = \frac{9!(10 \cdot 11 -10)}{9!} = 10 \cdot 11 - 10 =\boxed{\textbf{(B)}\;100.}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:01, 3 February 2016

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

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

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
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 10 Problems and Solutions

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