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# AMC 10 Problem

+3
118
2
+355

(I didn't understand the AoPS solution completely)

Using the letters \(A, M,O,S,\) and \(U,\) we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" \(USAMO\) occupies position

\((A) 112\)

\((B) 113\)

\((C) 114\)

\((D) 115\)

\((E)116\)

Jan 4, 2020

#1
+42
+4

Hello! I like the letters of the problem :-).

Notice that "USAMO" is one of the last numbers in the five letter words. If we let A=1, M=2, O=3, S=4, and U=5, we can write the numbers in term of "12345." Clearly, USAMO is only of the last letters in the sequence, as it starts with U. Note that "USAMO" is equal to 54123. Now, we can simply count backwords! The last number is 54321, followed by 54312, 54231, 54213,54132, and 54123. Thus, we can clearly see that "USAMO" is 5 away from the last number. There are 5! ways to order the numbers. The 5th to the last is 120-5= 115.

Jan 4, 2020
#2
+22078
+1

There are 5! possible 'words'n  (120)

AFTER usamo   is usaom   usmao   usmoa   usoam  usoma

115         116       117         118        119      120

Jan 4, 2020