1993 AIME Problems/Problem 1

Problem

How many even integers between 4000 and 7000 have four different digits?

Solution

The thousands digit is $\in \{4,5,6\}$ . If the thousands digit is even ( $4,\ 6$ , 2 possibilities), then there are only $\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \cdot 4 \cdot 8 \cdot 7 = 448$ .

If the thousands digit is odd ( $5$ , one possibility), then there are $5$ choices for the units digit, with $8$ digits for the hundreds and $7$ for the tens place. This gives $1 \cdot 5 \cdot 8 \cdot 7 = 280$ possibilities. Together, the solution is $448 + 280 = \boxed{728}$ .

1993 AIME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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1993 AIME Problems/Problem 1

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