1995 AJHSME Problems/Problem 23

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 $5*5*8*7=1400 \text{(B)}$ numbers.

Page

Toolbox

Search

1995 AJHSME Problems/Problem 23

Problem

Solution