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?

  • $\begingroup$ 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. $\endgroup$ May 4, 2011 at 3:19

2 Answers 2


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 $= 6!/(3!2!) \times 5!/(3!2!) \times \binom{7}{5}$.

  • $\begingroup$ How do you know that there are 7 spaces for the 5 vowels to go in? $\endgroup$
    – Krysten
    May 4, 2011 at 3:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – svenkatr
    May 4, 2011 at 3:42

(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.C.CC. (count: 10)
.CC.CC.C.C. (start double consecutive C pair)
.C.C.CC.CC. (count: 6)
.CCC.C.C.C. (start triple C string)
.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.

  • $\begingroup$ I forgot one possible combination: C1 C2 V1 C3 V2 C4 V3 C5 V4 C6 V5. I went back and edited the post, bringing the total up to (17)6!5!. $\endgroup$
    – Michael
    May 4, 2011 at 3:48
  • $\begingroup$ @Michael - your starting count of $6!5!$ is wrong because there are repeating vowels and consonants. Also, I don't see how you can get a factor of 17. I suspect you have made a mistake when you manually count the possible arrangements. $\endgroup$
    – svenkatr
    May 4, 2011 at 3:53
  • $\begingroup$ @Michael - That's a very interesting way to look at it. I guess if all else fails you can just look at all the different possibilities. However, I believe the answer is 12,600. $\endgroup$
    – Krysten
    May 4, 2011 at 3:55
  • $\begingroup$ @svenkar I made a prettier version of it, showing all 17 formats that the vowels and consonants can follow. If you can find anything wrong with the formats, please explain! $\endgroup$
    – Michael
    May 4, 2011 at 4:07
  • $\begingroup$ @Krysten I've noticed a problem! The subject/title of this question is "How many permutations of ENGINEERING do not contain consecutive vowels," which is the question that I answered. This, however, does not match up with the question that you ask in the body of your post, which asks, "how would I determine how many ways vowels can be ordered so that they are not next to each other?," which I find to be both ambiguous and quite different than the subject/title. $\endgroup$
    – Michael
    May 4, 2011 at 4:08

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