1995 AJHSME Problems/Problem 23

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Problem

How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?

$\text{(A)}\ 1120 \qquad \text{(B)}\ 1400 \qquad \text{(C)}\ 1800 \qquad \text{(D)}\ 2025 \qquad \text{(E)}\ 2500$

Solution

Count from left to right: There are 5 choices for the first digit, 5 choices for the second, 8 remaining choices for the third, and 7 remaining for the fourth, so there are $5*5*8*7= \boxed{\text{(B)}\ 1400}$ numbers.

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AJHSME/AMC 8 Problems and Solutions
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