# How many arrangements of MATHEMATICS are there in which each consonant is adjacent to a vowel?

How many arrangements of MATHEMATICS are there in which each consonant is adjacent to a vowel?

Now I have calculated it to be $$11 \cdot \frac{7!}{2!2!} \cdot \frac{4!}{2!}$$
for 11 distinct distributions of consonants among vowels with repetitions making sure that no two vowels are together. (If not, then at least one consonant will be not adjacent to any of the four vowels.)

I'm not sure about it and I would like to verify my answer. Any help will be appreciated.

• I don't get the 11, I get it to be $5{7!\over2!2!}{4!\over2!}$. The number of distributions of consonants and vowels seem to be 5 (with one more consonant it would be only one: cwccwccwccwc, then there's only 5 distinct ways to remove one). Commented Sep 7, 2015 at 10:05
• Oh k. I think I've included repetitive cases then.In these 5 cases, are the cases with vowels starting or ending the word considered? Commented Sep 7, 2015 at 10:15
• thanks @skyking i got my mistake. You have given a better way to explain though :) Commented Sep 7, 2015 at 10:27

With eight consonants there's only one way to distribute consonants and vowels (cvccvccvccvc) and removing one of them can only be done in five distinct ways (with vowels denoted by - for better visual appearance):

1. -cc-cc-cc-c

2. c-c-cc-cc-c

3. c-cc-c-cc-c

4. c-cc-cc-c-c

5. c-cc-cc-cc-

Then it's only to place the consonants in $${7!\over2!2!}$$ ways (since the order of m's and t's doesn't matter) and vowels in $${4!\over2!}$$ ways (since the order of a's doesn't matter).

The answer is: $$5{7!\over2!2!}{4!\over2!}$$

You can visualize each vowel as having two arms to which consonants can be attached

$-V-\;\;-V-\;\;-V-\;\;-V-$

There are 8 such arms,and for placing 7 consonants, 1 arm to be left, but if you leave out an arm between two vowels, one of the arms "disappears", so in effect there are only 5 ways of leaving out an arm,

$$\text{ thus answer} = 5 \cdot \frac{7!}{2!2!} \cdot \frac{4!}{2!}$$

ps

If $c$ is the # of consonants, and $v$ the # of vowels, the coefficient of 5 can be obtained as

$$\sum_{k=c-v}^{v-1}\binom{v-1}k\binom{v+1-k}{c-2k}$$

• But placing one at to the right of the first and none at the left of the second is the same as placing one to the left of the second and none after the first. Commented Sep 7, 2015 at 10:28
• @true blue anil yeah but both arms need not be occupied at the same time with consonants. That's how you get arrangements starting from vowels themselves. Commented Sep 7, 2015 at 10:32
• Your updated answer still doesn't seem correct. Besides it looks like you've picked "6" out of nowhere (without motivation). Commented Sep 7, 2015 at 10:40
• You are right, I was in a hurry, I have edited. Commented Sep 7, 2015 at 10:49