# Thread: Permutations problem with Restrictions

1. ## Permutations problem with Restrictions

Hi, I'm stuck on this problem and I have no idea how to approach it. Can somebody help me?

Find the number of 4-letter words that can be formed from the letters in the word QUESTION if no two vowels can be together.

Thanks.

2. Originally Posted by arden
Hi, I'm stuck on this problem and I have no idea how to approach it. Can somebody help me?

Find the number of 4-letter words that can be formed from the letters in the word QUESTION if no two vowels can be together.

Thanks.
If each word is only 4 letters, then no word can have more than two vowels, and words with 2 vowels are limited in their permutations.

For this, we would need to find the total number of permutations that are possible if we had only 2 letter words containing only vowels. Then we would have to multiply this by the total number of permutations that can exist of two letter words that contain only consonants. Note also that such combinations of consonants and vowles have only 2 possible arrangments: (v)(c)(v)(c) or (c)(v)(c)(v). For this reason, we would multiply the above product by 2.

Next, words containing 1 vowel need to be done in a similar process: We would need to find the total number of permutations of 1 letter words containing only vowles. Then we would need to multiply this by the total number of permutations of 3 letter words containing only consonants. Notice that this combination of 3 consonants and 1 vowel, has for arrangments: (v)(c)(c)(c), (c)(v)(c)(c), (c)(c)(v)(c), (c)(c)(c)(v). Therefore, we would need to multiply the above result by 4.

Last, words containing no vowels are the easiest to compute. We need only find the total number of permutations of 4 letter words containing only consonants.

Finally, we add the totals from each step to find the total number of possibilities.

3. See this PDF file.

4. Hello, arden!

Find the number of 4-letter words that can be formed from the letters
in the word QUESTION if no two vowels can be together.

We have four vowels {E,O,P,U} and four consonants {N,Q,S,T}.

We note that a 4-letter word cannot contain 3 vowels or 4 vowels.

Hence, there are three cases to consider: 0 vowels, 1 vowel, 2 vowels.

0 vowels
The word is formed from the four consonants.
. . There are 4! = 24 words with no vowels.

1 vowel
We select one vowel: 4 ways
We select three consonants: C(4,3) = 4 ways
Then the four letters are permuted: 4! ways.
Hence, there are: 4 × 4 × 24 = 384 words with one vowel.

2 vowels
We select two vowels: C(4,2) = 6 ways
We select two consonants: C(4,2) = 6 ways
We permute the four letters: 4! ways
So there are: 6 × 6 × 24 = 864 words with two vowels and two consonants.

But some of these have adjacent vowels . . . How many?
Select two vowels: C(4,2) = 6 ways
Select two consonants: C(4,2) = 6 ways
Duct-tape the two vowels together.
The three "letters" can be permuted in 3! = 6 ways.
The two vowels can have two orders (e.g., EO or OE).
So there are: 6 x 6 x 6 x 2 = 432 words with the two vowels adjacent.

Hence, there are: 864 - 432 = 432 words with two non-adjacent vowels.

Therefore, there are: .24 + 384 + 432 .= .840 words with non-adjacent vowels.

5. Thanks alot guys, I really appreciate it.