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

Revision as of 13:42, 17 March 2023

What is the value of $\dfrac{11!-10!}{9!}$ ?

$\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132$

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

$\dfrac{11!-10!}{9!}$ consider 10 as n $\dfrac{(n+1)!-n!}{(n-1)!}$

~IceMatrix

~savannahsolver

2016 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

@@ Line 17: / Line 17: @@
 <math>\dfrac{11!-10!}{9!}</math>
 consider 10 as n
-consider the nurmetor only
+<math>\dfrac{(n+1)!-n!}{(n-1)!}</math>
-$\dfrac{11!-10!}
-(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==