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So here is our "word" AEIOU. Then we need to find how many permutations contain EA and UO.

Then how many contain AE and EI

and how many end with O.

I know how to figure out some of these problems but it seems to be different depending on where the letters are in the word, so I'm not sure how to do EA since it's not the same as AE.

Would AE just be $\frac {5!}{1!1!1!1!}?$

Thanks guys, appreciate the help, I know there were a few similar problems but most seemed to be one letter only and not out of order from the original word.

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  • $\begingroup$ If it contains both EA and UO, you can just consider EA and UO to be single letters. You're permuting EA-I-UO, which is three "letters," so... $\endgroup$ – Akiva Weinberger Aug 10 '15 at 23:32
  • $\begingroup$ So is this similar to the finding AE, would it just be $\frac {5!}{1!1!1!1!}$ ? $\endgroup$ – androidguy Aug 10 '15 at 23:34
  • $\begingroup$ That's the number of permutations of five letters with no constraints. $\endgroup$ – Akiva Weinberger Aug 10 '15 at 23:35
  • $\begingroup$ How does this change with constraints? $\endgroup$ – androidguy Aug 10 '15 at 23:36
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Remember that $EA$ and $UO$ are being treated as single letters. Let $X = EA$ and $Y = UO$. How many permutations are there of $XIY$. The answer is $3!$.

Then how many contain AE and EI

This is equivalent to containing $AEI$. Treat that as a super letter. You have then two other letters. So how many permutations are there of three letters?

and how many end with O.

We have four slots and four distinct letters. $- - - - O$. How many ways are there to permute the four-letter prefix amongst four distinct characters? That's just $4!$.

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  • $\begingroup$ Okay, so for EA and UO, the answer is just $3!$, don't we have to do something like $\frac {3!}{1!1!1!1!}$ since we are declaring the number of letters in the original, or is this for a completely different type of problem? $\endgroup$ – androidguy Aug 10 '15 at 23:41
  • $\begingroup$ $1! = 1$. So that is implicit. If we have multiple occurrences of the same letter (say $n$ occurrences of a given letter), then we divide out by $n!$ to account for permutations of the same letter resulting in the same word. $\endgroup$ – ml0105 Aug 10 '15 at 23:42
  • $\begingroup$ Okay, that makes sense, so $a.$ and $b.$ are just $3!$ ? seems easier than it should be haha. $\endgroup$ – androidguy Aug 10 '15 at 23:44
  • $\begingroup$ That is correct. :-) $\endgroup$ – ml0105 Aug 10 '15 at 23:45
  • $\begingroup$ Okay, thanks a lot! I'm just a bit surprised because these seem more complicated than that, but thanks again!! $\endgroup$ – androidguy Aug 10 '15 at 23:46
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For the first part: How many ways are there to reorder SIX? For each one, replace $S$ with $EA$ and $X$ with $UO$.

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