Magoosh数学求助。 题目中set不是应该是没有重复元素的集合吗

charlienzwUID：3720038

From the letters in MAGOOSH, we are going to make three-letter "words." Any set of three letters counts as a word, and different arrangements of the same three letters (such as "MAG" and "AGM") count as different words. How many different three-letter words can be made from the seven letters in MAGOOSH?
答案是135 给了两类情况，含有两个“O”的15种，一个O的120种。

可是说的是任意三个字母的set，set是集合的意思，无论是OG中还是我们所学的集合定义中都是没有重复元素的，所以为什么要考虑两个“O”的情况？

charlienzwUID：3720038

官方解答：
Let's break this into two cases, to deal with the double letter.

Case I: "words" with two O's

The non-O letter could take any of three spaces in the word:

XOO

OXO

OOX

In any of those scenarios, the X could be replaced by one of five letters: {M, A, G, S, H}. Five letters in three cases: Case I results in 15 "words" with two O's.

Case II: "words" with one O or no O's

Now, we can ignore the second O. We are simply picking 3 letters from the pool: {M,A,G,O,S,H}.

first letter of word = 6 possibilities

second letter of word = 5 possibilities

third letter of word = 4 possibilities

6*5*4 = 6*20 = 120 in Case II

Now, add the possibilities from the two cases.  

Total = 15 + 120 = 135

Orion_ChangUID：3582304

你没读懂题意。。。

题目的意思是问总共能组合出多少个不同的单词

而 AOO 和 AOO 不管在哪个宇宙都不可能算成两个不同的单词因为它们有完全一致的字母组合
（虽然它没有明确说明这句话，而是作为一个assumption）