# Probability help....?

Please help me !

Thanks !

### 1 Answer

- The GnosticLv 76 months agoFavorite Answer
Unfortunately, I have found ambiguity.

"no consonant falls between any two vowels" could mean two (or more) consonants could appear together between two vowels, or absolutely no consonants could appear between two vowels.

In the first case, VVCCV would be acceptable because each consonant is between a vowel and another consonant. In the second case, it would not be acceptable.

Proceeding, assuming case 1:

Acceptable 5-letter patterns: VVVCC, VVCCV, VVCCC, VCCVV, VCCVC, VCCCV, CVVVC, CVVCC, CVCCV, CCVVV, CCVVC, CCCVV.

Acceptable 6-letter patterns: VVVCCC, VVCCCV, VVCCVC, VCCCVV, VCCVVC, CVCCVV, CVVVCC, CVVCCV, CCVVVC, CCCVVV.

Total 5-letter words: (6!/(6-5)!) = 720

Total 6-letter words: (6!/(6-6)!) = 720

Each of the 5-letter patterns contains 3 of one group and two of the other group, so they would produce 3! * (3!/(3-2)!) = 6 * 6 = 36 words

Each of the 6-letter patterns contains 3 of each group, for a total of 6 * 6 = 36 words. (Gee, what a coincidence!)

There are 12 5-lettter patterns for a total of 432 words.

There are 10 6-letter patterns for a total of 360 words.

Therefore, the probability of getting an "acceptable" word is 792/1440 = 0.55

Now, assuming case 2:

Acceptable 5-letter patterns: VVVCC, VVCCC, CVVVC, CVVCC, CCVVV, CCVVC, CCCVV.

Acceptable 6-letter patterns: VVVCCC, CVVVCC, CCVVVC, CCCVVV.

There are 7 5-lettter patterns for a total of 252 words.

There are 4 6-letter patterns for a total of 144 words.

Therefore, the probability of getting an "acceptable" word is 396/1440 = 0.275

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