# 1983 AIME Problems/Problem 10

## Problem

The numbers $1447$, $1005$ and $1231$ have something in common: each is a $4$-digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there?

## Solution

### Solution 1

Suppose that the two identical digits are both $1$. Since the thousands digit must be $1$, only one of the other three digits can be $1$. This means the possible forms for the number are

$11xy,\qquad 1x1y,\qquad1xy1$

Because the number must have exactly two identical digits, $x\neq y$, $x\neq1$, and $y\neq1$. Hence, there are $3\cdot9\cdot8=216$ numbers of this form.

Now suppose that the two identical digits are not $1$. Reasoning similarly to before, we have 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 answer is $216+216=\boxed{432}$.

### Solution 2

Consider a sequence of $4$ digits instead of a $4$-digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\frac{1}{10}$. This means we can find all possible sequences with one digit repeated twice, and then divide by $10$.

If we let the three distinct digits of the sequence be $a, b,$ and $c$, with $a$ repeated twice, we can make a table with all possible sequences:

$$\begin{tabular}{ccc} aabc & abac & abca \\ baac & baca & \\ bcaa && \\ \end{tabular}$$

There are $6$ possible sequences.

Next, we can see how many ways we can pick $a$, $b$, and $c$. This is $10(9)(8) = 720$, because there are $10$ digits, from which we need to choose $3$ with regard to order. This means there are $720(6) = 4320$ sequences of length $4$ with one digit repeated. We divide by 10 to get $\boxed{432}$ as our answer.