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2015 AMC 8 Problems/Problem 10

Revision as of 19:49, 12 May 2023 by Thestudyofeverything (talk | contribs) (→‎Solution)

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 (HOW TO THINK CRITICALLY!!!)

https://youtu.be/-VcuRyDB_wI

~Education, the Study of Everything

Video solution

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

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

https://youtu.be/2nfFg8JXKFE

~savannahsolver

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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

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