2015 AMC 8 Problems/Problem 10

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 1

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.


If you're too lazy to do $9 \times 9 \times 8 \times 7= 4536$, we can notice $9 \times 9 = 81$ and $8 \times 7=56$ The unit digit must be 6. Therefore the answer is the only answer with a a unit digit of 6. $\boxed{B}$)

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

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