1986 AMC 12 Problem 10

Question

The

120

permutations of AHSME are arranged in dictionary order, as if each were an ordinary five-letter word. The last letter of the

86

th word in this list is

A

H

S

M

E

Group the words into blocks of

4!

according to their first letter

Within the

M

-block, group by the second letter and then list only the needed small block

Each first letter occupies a block of

4!=24

words. Positions

73

through

96

begin with

M,

so the

86

th word is the

14

th word in that block. The first

6

begin with

MA,

the next

6

with

ME,

and the

13

th through

18

th begin with

MH.

The first two of those are

MHAES

and

MHASE.

Thus the

86

th word ends in E.

Therefore the correct answer is E.

Problem 10 in other years