"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?
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...
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.