# Permutation Simple Question

• July 15th 2008, 11:44 PM
mander
Permutation Simple Question
"How many permutations exist of the letters in the word COMPUTER?"

I said 8! because there are 8 letters and so it should just be n! where n is the number of choices. Right?

"How many permutations end in a vowel?"

I thought 7!3! but I'm not certain. Because there are 3 vowels and I want one of them in the last position, and the other 7 letters can be whatever else is left..

Is this correct? If not, can someone tell me what I am doing wrong? (Nerd)
• July 16th 2008, 12:08 AM
CaptainBlack
Quote:

Originally Posted by mander
"How many permutations exist of the letters in the word COMPUTER?"

I said 8! because there are 8 letters and so it should just be n! where n is the number of choices. Right?

"How many permutations end in a vowel?"

I thought 7!3! but I'm not certain. Because there are 3 vowels and I want one of them in the last position, and the other 7 letters can be whatever else is left..

Is this correct? If not, can someone tell me what I am doing wrong? (Nerd)

How many end in E?
How many end in U?
How many end in O?

How many does that make?

RonL
• July 16th 2008, 12:23 AM
mander
I don't know how to figure that out.
• July 16th 2008, 12:27 AM
mr fantastic
Quote:

Originally Posted by mander
I don't know how to figure that out.

If it ends in E how many letters are there in front of the E that can re-arrange .....? etc.
• July 16th 2008, 12:30 AM
mander
7... Or is this a trick question? I already said 7 though..
• July 16th 2008, 12:32 AM
mr fantastic
Quote:

Originally Posted by mander
7... Or is this a trick question? I already said 7 though..

How many end in E + How many end in U + How many end in O $\neq7! \, 3! \, .......$
• July 16th 2008, 12:38 AM
mander
Yes, but I implied there were 7 choices before the vowel and was wondering about the vowel part specifically, which is why I said 7! originally... it was the 3! that wasn't sitting well with me. There's a miscommunication here, and I'm sorry for that.

anyway, I am going with 7! * C(3,1) ... I will check with someone in math lab tomorrow. Sorry for wasting your time...
• July 16th 2008, 12:41 AM
mr fantastic
Quote:

Originally Posted by mander
Yes, but I implied there were 7 choices before the vowel and was wondering about the vowel part specifically, which is why I said 7! originally... it was the 3! that wasn't sitting well with me. There's a miscommunication here, and I'm sorry for that.

anyway, I am going with 7! * C(3,1) ... I will check with someone in math lab tomorrow. Sorry for wasting your time...

How many end in E: 7!
How many end in U: 7!
How many end in O: 7!

Total: 7! + 7! + 7! = 3*7!

This is actually the same as 7! * C(3,1) but I'm not sure that you got the correct answer for the correct reason.
• July 16th 2008, 12:43 AM
Twig
hi
Hi!

Isnt this a combination of "k-permutations" and the addition principle.

How many end in E?
If we subtract the letter E from the rest of the letters we have 7 left.
And seven letters can be permutated in 7! ways.

The same goes for O and U.

So 7! + 7! + 7! = 15120 ways of permutating the word COMPUTER when the word ends with a vowel.

If the question was "How many ways can the word be permutated and DONT end with a vowel?" You would just take the total permutations 8! = 40320
and subtract 15120

40320 - 15120 = 25 200

I HOPE this is right, I am very very new at discrete math though.
Plz correct me if Im wrong.
• July 16th 2008, 03:52 AM
CaptainBlack
Quote:

Originally Posted by mr fantastic
How many end in E: 7!
How many end in U: 7!
How many end in O: 7!

Total: 7! + 7! + 7! = 3*7!

This is actually the same as 7! * C(3,1) but I'm not sure that you got the correct answer for the correct reason.

Amazing that's what I was just thinking(Giggle)

RonL