[Math] How many permutations of the letters AEIOU contain the strings EA and UO

combinatoricsdiscrete mathematicspermutations

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.

Best Answer

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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