Choose 5 places out of 11 11C5 Only one way to arrange aeiou And the rest 6 can be arranged in any way possible There are two 't' es 6! / 2! =>11C5 * 360 =166320
Yeah, that's how I did it. It's like, if you choose 5 from 11, they have to be arranged in a particular way (in the exact order a, e, i, o, u) so this would be a combination, not a permutation since there is only 1 way this is true. And the rest can go where ever so 6! and obviously divide by the overcount of 2! for 166,320. It's interesting how diverse these problems can get.
Pretty late comment, I understand this method and I know why it works. Because out of the 11 spots, we choose 5 of them to fill up the vowels (A,E,I,O,U) in that order. And hence the other letters can be arranged in 6!/2! ways. But, what if you had reoccuring a’s or e’s? Like say we had PERMUTAATIOON ( 2 extra O’s and A’s). How would you address this problem?
We can also select 5 space from 11 space for vowel as 11C5 and then we can arrange rest of letter as 6!/2!. So answer will be equal to 11C5*(6!/2!) or 166320. Note 11C5= 11!/(5!(11-5)!=11!/(5!6!) Your regards Syed Azhad Hussain.
We have the letters P E R M U T A T I O N, and the question is asking how many words we can form with those letters with the vowels ( A, E, I, O, U) in that order.
