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.

Solution 3 (Complementary Counting)

We'll use complementary counting. We will split up into $3$ cases: (1) no number is repeated, (2) $2$ numbers are repeated, and $2$ other numbers are repeated, (3) $3$ numbers are repeated, or (4) $4$ numbers are repeated.

Case 1: There are $9$ choices for the hundreds digit (it cannot be $1$ ), $8$ choices for the tens digit (it cannot be $1$ or what was chosen for the hundreds digit), and $7$ for the units digit. This is a total of $9\cdot8\cdot7=504$ numbers.

Case 2: One of the three numbers must be $1$ , and the other two numbers must be the same number, but cannot be $1$ (That will be dealt with in case 4). There are $3$ choices to put the $1$ , and there are $9$ choices (not $1$ ) to pick the other number that is repeated, so a total of $3\cdot9=27$ numbers.

Case 3: We will split it into $2$ subcases: one where $1$ is repeated $3$ times, and one where another number is repeated $3$ times. When $1$ is repeated $3$ times, then one of the digits is not $1$ . There are $9$ choices for that number, and $3$ choices for its location,so a total of $9\cdot3=27$ numbers. When a number other than $1$ is repeated $3$ times, then there are $9$ choices for the number, and you don't have any choices on where to put that number. So in Case 3 there are $27+9=36$ numbers

Case 4: There is only $1$ number: $1111$ .

There are a total of $1000$ $4$ -digit numbers that begin with $1$ (from $1000$ to $1999$ ), so by complementary counting you get $1000-(504+27+36+1)=\boxed{432}$ numbers.

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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