Arnold F. answered • 04/28/16
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Here's a hint. If the question were just how many permutations are there with no restrictions the answer would be 10!. Since you want to restrict the vowels to be in alphabetical order you have to divide this result by 5!.
Can you explain why this works?
Arnold F.
Not sure how you're getting that. It would be just 10!/5! .
The reasoning for this would be as if you had 10 letters and five of those letters were all A's.
The calculation above would be the number of permutations of these 10 letters.
then take the second A replace it with E take the third A replace it with I take the fourth A replace it with O take the fifth a replace it with U.
this would give you a permutation with the vowels in the correct order.
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04/28/16
Bianca J.
Ohh that makes sense. Thank you so much! I don't know why I thought permutation would play a role in this.
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04/28/16
Joseph G.
15120 is correct. Abstemious has 2 S’s and therefore 10!/5! Will overcount by a factor of 2!
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04/05/19
Bianca J.
04/28/16