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?

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

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 then $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\textbf{(D) } 8}$

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

Note: for each combination of 7 numbers, exactly 1 is in increasing order -Williamgolly

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 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
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2005 AMC 10B Problems/Problem 18

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