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2005 AMC 10B Problem 18

2005 AMC 10B Problem 18AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 3 of 5 · about 2 min · position P16-20

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?

A)
$1$
B)
$2$
C)
$7$
D)
$8$
E)
$9$

Answer

Key insight

An increasing string of 7 distinct digits is determined by which 7 of the 8 allowed digits are used, i.e. which single digit is left out.

Solution

The allowed digits are $2, 3, 4, 5, 6, 7, 8, 9$ : eight of them. The seven positions $a$ through $g$ must be filled with seven distinct allowed digits.

Once we decide which seven digits to use, the increasing-order requirement leaves only one way to place them. So a telephone number corresponds exactly to a choice of $7$ digits out of $8$ , or equivalently to the one digit that is omitted.

There are $8$ digits that could be the omitted one, so there are $8$ telephone numbers.

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Why this works

"Distinct and in increasing order" is a signal that order carries no information: the set of digits determines the string. Counting sets instead of strings turns a permutation-flavored problem into a combination, and $\binom{8}{7} = \binom{8}{1}$ is easiest read as "pick the one left out."

The trap

Counting arrangements of the digits (permutations) when the increasing-order condition fixes the arrangement completely.

Common mistakes

Counting arrangements of the digits (permutations) when the increasing-order condition fixes the arrangement completely.
Miscounting the allowed digits as $9$ or $7$ (forgetting that both $0$ and $1$ are excluded, or dropping $9$ ).

Techniques

Map the objects to something easier to count