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The word is "ENGINEERING".

The number of ways that the consonants can be ordered is 6! / 3!2!

The number of ways that the vowels can be ordered is 5! / 3!2!

But how would I determine how many ways vowels can be ordered so that they are not next to each other?

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    Think of using the vowels as separators, and thus you'll have 6 bins to place 6 consonants. Then you need to place a consonant between each spacer, and you'll be left with 2 consonants to place in any of the 6 bins. And of course take into account the double letters.2011-05-04

2 Answers 2

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Imagine that you arrange the consonants first. There are six consonants which you can arrange in $6!/(3!2!)$ ways.

Now there are 7 spaces for the 5 vowels to go into but only one vowel can go into each space. So you choose 5 of the 7 available spaces and put a permutation of the vowels into these spaces.

Total number of arrangements with no consectutive vowels $= 6!/(3!2!) \times 5!/(3!2!) \times \binom{7}{5}$.

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    @krysten - You know that the consonants have to separate the vowels. Counting one space in front of the consonants, one at the end and the five in the middle gives us 7 seven spaces. For example, look at .N.G.N.R.N.G. which is an arrangement of consonants with the dots representing places where vowels can go. There are seven dots, which are the seven places where a vowel can go into.2011-05-04
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(edit: Answer has been updated to include four missed combinations and the fact that permutations are not considered unique if their spelling matches that of another permutation.)

12345678901  (11 letters in ENGINEERING) C.C.C.C.C.C (alternating sequence, bounded on both sides by C) .C.C.C.C.CC  (start single consecutive C pair, one end bounded by C) .C.C.C.CC.C .C.C.CC.C.C .C.CC.C.C.C .CC.C.C.C.C CC.C.C.C.C. C.CC.C.C.C. C.C.CC.C.C. C.C.C.CC.C. C.C.C.C.CC. (count: 10) .CC.CC.C.C. (start double consecutive C pair) .CC.C.CC.C. .CC.C.C.CC. .C.CC.CC.C. .C.CC.C.CC. .C.C.CC.CC. (count: 6) .CCC.C.C.C. (start triple C string) .C.CCC.C.C. .C.C.CCC.C. .C.C.C.CCC. (count: 4) 

In all of these formats, there are 6!/(3!2!) ways to order the C's (i.e. consonants) and 5!/(3!2!) ways to order the .'s (i.e. vowels,) so there should be (21)(6!/3!2!)(5!/3!2!), or 12,600, permutations in which there are no adjacent vowels.

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    The question didn't state that all of the E's would be considered identical. She did indeed say "6!/3!2!" and "5!/3!2!," but it didn't click in my mind as to what she was implying. Thanks for clearing that up. I apparently did indeed leave out those combinations, and yep... that would result in the answer matching the back of the book: (21)6!5!/((3!2!)^2) = 12,600. I'll edit the answer to reflect this... sorry for missing the 3!2!s and the four extra combinations. :)2011-05-04