Difference between revisions of "2015 AMC 8 Problems/Problem 10"

(Video solution)
(Video solution)
 
Line 12: Line 12:
  
 
https://www.youtube.com/watch?v=OESYIYjZFdk
 
https://www.youtube.com/watch?v=OESYIYjZFdk
 +
 +
https://youtu.be/2nfFg8JXKFE
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2015|num-b=9|num-a=11}}
 
{{AMC8 box|year=2015|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:34, 30 March 2022

Problem

How many integers between $1000$ and $9999$ have four distinct digits?

$\textbf{(A) }3024\qquad\textbf{(B) }4536\qquad\textbf{(C) }5040\qquad\textbf{(D) }6480\qquad \textbf{(E) }6561$

Solution

 There are $9$ choices for the first number, since it cannot be $0$, there are only $9$ choices left for the second number since it must differ from the first, $8$ choices for the third number, since it must differ from the first two, and $7$ choices for the fourth number, since it must differ from all three.  This means there are $9 \times 9 \times 8 \times 7=\boxed{\textbf{(B) }4536}$ integers between $1000$ and $9999$ with four distinct digits.

Video solution

https://youtu.be/Zhsb5lv6jCI?t=272

https://www.youtube.com/watch?v=OESYIYjZFdk

https://youtu.be/2nfFg8JXKFE

~savannahsolver

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS