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

(Solution 2)
(Solution 3)
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We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!}</cmath> which will get us <math>110 -10 = \boxed{\textbf{(B)}\;100.}</math>
 
We can use subtraction of fractions to get <cmath>\frac{11!-10!}{9!} = \frac{11!}{9!} - \frac{10!}{9!}</cmath> which will get us <math>110 -10 = \boxed{\textbf{(B)}\;100.}</math>
 
==Solution 3==
 
==Solution 3==
 +
 +
Factoring out <math>9!</math>
  
 
==See Also==
 
==See Also==

Revision as of 14:14, 5 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 1

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


Solution 2

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

Solution 3

Factoring out $9!$

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
2016 AMC 12A (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 12 Problems and Solutions

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