View source for 2016 AMC 10A Problems/Problem 1

==Problem==

What is the value of <math>\dfrac{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 1==

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

==Solution 2==

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


==Solution 3==

consider 10 as n 
consider the nurmetor only 
(n+1)!-n!=(n+1)n!+(-1)n!
= n(n!)

now with the deniminator 

n(n!)/(n-1)!

n(n-1)!=n!

so 
= n(n(n-1)!)/(n-1)!
divide the (n-1)!
we get n time n 
= n^2
= 10^2
= 100

==Video Solution==
https://youtu.be/VIt6LnkV4_w

~IceMatrix

https://youtu.be/CrS7oHDrvP8

~savannahsolver

==See Also==
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}
{{AMC12 box|year=2016|ab=A|before=First Problem|num-a=2}}
{{MAA Notice}}