1990 AJHSME Problems/Problem 12

Problem

There are twenty-four $4$-digit numbers that use each of the four digits $2$, $4$, $5$, and $7$ exactly once. Listed in numerical order from smallest to largest, the number in the $17\text{th}$ position in the list is

$\text{(A)}\ 4527 \qquad \text{(B)}\ 5724 \qquad \text{(C)}\ 5742 \qquad \text{(D)}\ 7245 \qquad \text{(E)}\ 7524$

Solution

For each choice of the thousands digit, there are $6$ numbers with that as the thousands digit. Thus, the six smallest are in the two thousands, the next six are in the four thousands, and then we need $5$ more numbers.

We can just list from here: $5247,5274,5427,5472,5724 \rightarrow \boxed{\text{B}}$.

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png