AIME 1983 P10

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Problem Statement

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?

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Solution

$$\text{Fix the first digit to be 1}$$ $$\text{Case 1: Repeated 1s: One of the 3 remaining digits must be 1. There are } C(9, 2) \text{ ways to pick the other two digits and 3! ways to rearrange the digits} $$ $$\implies {9\choose 2} \times 3! = 216 \text{ ways}$$ $\text{Case 2: Not Repeated 1s: None of the remaining digits can be 1. There are P(9, 2) ways to pick the digits}$$ $$\text{ and } \dfrac{3!}{2!} \text{ ways to rearrange the digits}$$ $$\implies 9 \times 8 \times \dfrac{3!}{2!} = 216 \text{ ways}$$ $$\therefore \text{there are } 432 \text{ numbers in total}$$