1986 AHSME Problems/Problem 10

Problem

The permutations of are arranged in dictionary order as if each were an ordinary five-letter word. The last letter of the th word in this list is:

Solution 1

We could list out all of the possible combinations in dictionary order.

MAEHS MAESH MAHES MAHSE MASEH MASHE MEAHS MEASH MEHAS MEHSA MESAH MESHA MHAES MHASE

We find that the combination ends with the letter E. So the answer is .

Solution 2

We can do this problem without having to list out every single combination. There are distinct letters, so therefore there are ways to rearrange the letters. We can divide the different combinations into 5 groups. Words that start with , words that start with and so on... Combinations - start with , combinations - start with , combinations - start with , combinations - start with , and combinations - start with . We are only concerned with combination , so we focus on combinations -. We can divide the remaining 24 combinations into 4 groups of 6, based upon the second letter. Combinations - begin with , combinations - begin with , combinations - begin with , and combinations - begin with . Combination begins with . Now we can fill in the rest of the letters in alphabetical order and get (as is ). The last letter of the word is , so the answer is .

See also

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