# Difference between revisions of "2005 AMC 10B Problems/Problem 18"

## Problem

All of David's telephone numbers have the form $555-abc-defg$, where $a$, $b$, $c$, $d$, $e$, $f$, and $g$ are distinct digits and in increasing order, and none is either $0$ or $1$. How many different telephone numbers can David have?

$\mathrm{(A)} 1 \qquad \mathrm{(B)} 2 \qquad \mathrm{(C)} 7 \qquad \mathrm{(D)} 8 \qquad \mathrm{(E)} 9$

## Solution

The only digits available to use in the phone number are $2$, $3$, $4$, $5$, $6$, $7$, $8$, and $9$. There are only $7$ spots left among the $8$ numbers, so we need to find the number of ways to choose $7$ numbers from $8$. The answer is just $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\mathrm{(D)}\ 8}$

Alternatively, we could just choose $1$ out of the $8$ numbers to not be used. There are obviously $\boxed{8}$ ways to do so.

## See Also

 2005 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 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

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