2015 AMC 8 Problems/Problem 10

Revision as of 10:36, 29 December 2015 by Usa (talk | contribs) (Solution 1)

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

The question can be rephrased to "How many four-digit positive integers have four distinct digits," since numbers between $1000$ and $9999$ are four-digit integers. There are $9$ choices for the first number, since it cannot be $0$, There are only $9$ choices 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.

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