Originally Posted by

**Jame** Hello all, I am trying to count the number of arrangements of the word MATHEMATICS where each consonant in adjacent in a vowel. (7 consonants, 4 vowels)

Consider the following diagram:

$\displaystyle \left(\;\;\;\;\right)\;\;v\;\;\left(\;\;\;\;\right )\;\;v\;\;\left(\;\;\;\;\right)\;\;v\;\;\left(\;\; \;\;\right)\;\;v\;\;\left(\;\;\;\;\right)$

where v is a vowel and $\displaystyle \left(\;\;\;\;\right)$ is a bubble for consonants to go in

Now in order to deal with the constraint of each consonant being next to a vowel, I believe the following is true

Two of the three middle bubbles must contain exactly two consonants

One of the end bubbles must contain exactly one consonant.

So we have the following possible unordered distributions (partitions?) of 7:

{2,2,1,1,1} and {2,2,2,1,0}

For {2,2,2,1,0}, all three twos must go in the middle three bubbles. The 0,1 can go on either end.

This leads to two possible distribution of consonants 0v2v2v2v1 and 1v2v2v2v0

For {2,2,1,1,1}, choose two of the three middle bubbles for the 2's, fill in the rest with 1's.

This leads to three possible distribution of consonants 1v2v2v1v1, 1v2v1v2v1, 1v1v2v2v1

So there are 5 possible ways we can place the vowels and consonants. Now we just need to order them

$\displaystyle \frac{7!}{2!2!}$ ways to order consonants

$\displaystyle \frac{4!}{2!}$ ways to order vowels

Final answer:

$\displaystyle 5 * \frac{7!}{2!2!} * \frac{4!}{2!}$

Any thoughts? clarifications? glaring errors?

Thank you for your help.