1983 AIME Problems/Problem 10

Problem

The numbers $1447$ , $1005$ , and $1231$ have something in common. Each is a four-digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

Solution

Suppose the two identical digits are both one. Since the thousands digits must be one, the other one can be in only one of three digits,

$11xy,\qquad 1x1y,\qquad1xy1,\qquad11yx,\qquad1y1x,\qquad1yx1$

Because the number must have exactly two identical digits, $x\neq y$ , $x\neq1$ , and $y\neq1$ . Hence, there are $6\cdot9\cdot8=432$ numbers of this form.

Suppose the two identical digits are not one. Therefore, consider the following possibilities,

$1xxy,\qquad1xyx,\qquad1yxx.$

Again, $x\neq y$ , $x\neq 1$ , and $y\neq 1$ . There are $3\cdot9\cdot8=216$ numbers of this form.

Thus, the desired answer is $432+216=\boxed{648}$ .

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1983 AIME Problems/Problem 10

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