Figuring out simple arrangements?

Question

I've been studying questions related to permutations and combinations, and this type of question always comes up without much explanation from my books/teachers.

For example, let's say you have 7 letters total, 5 of those Xs and 2 of those Os. (XXXXXOO)
How many unique arrangements (7 long) can you make with the letters?
But the question I really have is, how would you solve this mathematically if you changed all the amounts?  That is, what is the trick to solving this sort of problem?

M3 · Accepted Answer

there is no "trick", a standard formula for dealing with such cases.
if all items were distinct, it'd be 7!
since there are 5 of one kind & 2 of another, divide by 5!2!, ie
 7!
----- .  
5!2!

in general, n!/(n1!n2!...) where n1 are of one kind, n2 of another, ....

http://mathematics.learnhub.com/lesson/5607-permutations-in-which-some-are-alike

Tim · Answer

So the idea is that if you had 7 different letters, there would be 7! permutations you can make.  So we start with that as the numerator.  However, this overcounts in this case, since, for example, there are 5 X's, and if you take any given word, like XXXXXOO, and rearrange the X's, it's the same word.  So we divide by 5!, the number of ways we could rearrange just the X's, since we're counting 5! as many words as we should since for each word, we also count all the rearrangements of the X's, and also divide by 2!, the number of ways we could rearrange just the O's.

Final answer - 7!/(5!*2!)

Trending News

Figuring out simple arrangements?

2 Answers