# Arrangements of the word MATHEMATICS with restriction.

• Feb 16th 2013, 11:31 AM
Jame
Arrangements of the word MATHEMATICS with restriction.
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

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

Any thoughts? clarifications? glaring errors?

• Feb 16th 2013, 02:04 PM
Plato
Re: Arrangements of the word MATHEMATICS with restriction.
Quote:

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

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

Any thoughts? clarifications? glaring errors?

I did it an entirely different ways getting the same answer.
• Feb 16th 2013, 02:24 PM
Jame
Re: Arrangements of the word MATHEMATICS with restriction.
Thank you very much for replying! May I ask how you arrived at your answer? I am always interested in seeing a different way to do a problem.
• Feb 16th 2013, 02:38 PM
Plato
Re: Arrangements of the word MATHEMATICS with restriction.
Quote:

Originally Posted by Jame
Thank you very much for replying! May I ask how you arrived at your answer? I am always interested in seeing a different way to do a problem.

I did it using a generating function.
Expand $\displaystyle \left( {1 + x} \right)^2 \left( {1 + x + x^2 } \right)^3$.
5 is the coefficient of $\displaystyle x^7$. Then use your arrangement.
• Feb 16th 2013, 03:11 PM
Jame
Re: Arrangements of the word MATHEMATICS with restriction.
Interesting. So we build this function to partition a number into five summands, two of which are between 0 and 1 and three of which are between 0 and 2.

And as it turns out, at least two of the five summands must be 2 in order to get a sum of 7. (the ones mentioned eariler)

Thanks again!