Problem with Alphabet

[lladams]

How many ways are there to arrange the 26 letters of the alphabet so that no pair of vowels appear consecutively (Y is considered a constant)?

 

[Count Iblis]

How many ways are there to arrange the 26 letters of the alphabet so that no pair of vowels appear consecutively (Y is considered a constant)?

Suppose you have 26 positions for the letters. You choose positions for the vowels. You can put down the vowels in all possible ways by first choosing a particular ordering (5! possible ways) and then putting them down in that order from left to right. For the first vowel you then have 26 - 8 positions available. For the next you have two positions less available etc. etc. Once you have filled 5 positions with five vowels you have 21! ways of putting in the non-vowels.

 

[lladams]

i don't understand where you got 26-8 from, how is that possible positions left after the first position?

 

[pka]

i don't understand where you got 26-8 from

Well I don't either!
Remove the five vowels, then the 21 letters left act as separators. The separators create 22 positions the vowels can occupy. Thus $\displaystyle {22}\choose 5$ is the number of places to put the vowels. There are $\displaystyle 5!$ ways to arrange the vowels and $\displaystyle 21!$ ways to arrange the separators.

ANSWER: \(\displaystyle {{22}\choose 5}{\(5!\)}{\(21!\)}\)

 

[Count Iblis]

i don't understand where you got 26-8 from, how is that possible positions left after the first position?

Sorry, I made a mistake. You can evaluate it as follows:

$\displaystyle \sum_{i_{1}=1}^{26}\sum_{i_{2}=i_{1}+2}^{26}\sum_{i_{3}=i_{2}+2}^{26}\sum_{i_{4}=i_{3}+2}^{26}\sum_{i_{5}=i_{4}+2}^{26}1$

Here $\displaystyle i_{1}$ is the position of the first vowel. I let the sum go to 26 but if it is more than 18 it will yield zero because then there is no room left to oput down the other vowels. This summation must be multiplied by 5! to account for all the possible orders in which you can put them down.

 

[pka]

Count Iblis, I cannot imagine where you studied your combinatorics.
What a mess you have posted.
Read my response to this problem.
Bertram Russell said: “If it is meaningful it can be said simply.”
Paul Erdos says the same applies counting problems.

 

[Count Iblis]

Count Iblis, I cannot imagine where you studied your combinatorics.
What a mess you have posted.
Read my response to this problem.
Bertram Russell said: “If it is meaningful it can be said simply.”
Paul Erdos says the same applies counting problems.

I don't understand it. How can there be 22 positions for the five vowels? I thought that no two vowels can be next to each other. That problem is practically the same as the problem of the one dimensional hard sphere gas. The solution I wrote down is the analogue of the partition function of that problem, which is not so difficult to evaluate by the way.

 

[pka]

I don't understand it. How can there be 22 positions for the five vowels? I thought that no two vowels can be next to each other. That problem is practically the same as the problem of the one dimensional hard sphere gas. The solution I wrote down is the analogue of the partition function of that problem, which is not so difficult to evaluate by the way.

No it is not. Remember that a ‘a little knowledge (i.e. understanding) is a dangerous thing.” If you really cannot follow the solution then you are a long way from understanding general counting solutions.

 

[Denis]

I don't understand it. How can there be 22 positions for the five vowels?

Simplify it to 3 consonants :
vcvcvcv : 1 more position than consonants

Same with full alphabet, of course: 21 consonants, so 22 positions.

 

[Count Iblis]

I don't understand it. How can there be 22 positions for the five vowels? I thought that no two vowels can be next to each other. That problem is practically the same as the problem of the one dimensional hard sphere gas. The solution I wrote down is the analogue of the partition function of that problem, which is not so difficult to evaluate by the way.

No it is not.

Yes it is (the number of ways you can place the five vowels down is exactly a discrete version of the hard spere problem).

But anyway, we all know that people no matter how knowlegable can sometimes fail to see very simple things. When that happens it is extremely rude to use that to make the ridiculous remarks as you just did.

If you really cannot follow the solution then you are a long way from understanding general counting solutions

Wrong, because I'm an expert in statistical mechanics and I couldn't follow your solution. So, that disproves your ridiculous thesis.

 

[Count Iblis]

I don't understand it. How can there be 22 positions for the five vowels?

Simplify it to 3 consonants :
vcvcvcv : 1 more position than consonants

Same with full alphabet, of course: 21 consonants, so 22 positions.

Yes, I see this. But not all the 22 positions are really available. I guess I have to think about this method a little more...

 

[Count Iblis]

Ok, I see it now. There are 22 potential places of which 5 are used. This gives the Binomial[22,5] factor.

 

[Denis]

Yes, I see this. But not all the 22 positions are really available. I guess I have to think about this method a little more...

WHY are the 22 positions not really available :shock:
1b2c3d4f5g6h7j8k9l10m11n12p13q14r15s16t17v18w19x20y21z22
Doesn't that make it clear?

Edit: Thanks for agreeing :wink:

 

[Count Iblis]

Yes, I see this. But not all the 22 positions are really available. I guess I have to think about this method a little more...

WHY are the 22 positions not really available :shock:
1b2c3d4f5g6h7j8k9l10m11n12p13q14r15s16t17v18w19x20y21z22
Doesn't that make it clear?

Edit: Thanks for agreeing :wink:

What I didn't get at first is that you choose a particular relative ordering for the 21 consonants, but you don't fix the absolute positions. You then put in the five vowels and then all the absolute positions are fixed.

 

[Count Iblis]

Anyway, this is an interesting problem. Suppose we demand that the minimum distance between the vowels is d. I can then still solve the problem using my method, I'm interested if the simple method that pka gave can be modified to work too.

This is how I would solve it. Choose again a particular ordering for the five vowels. Then there are:

\(\displaystyle \L\sum_{i_{1}=1}^{26}\L\sum_{i_{2}=i_{1}+d}^{26}\cdots\sum_{i_{5}=i_{4}+d}^{26} 1\)

ways to put in the vowels. We need to multiply this summation by the 5! possible orderings of the vowels and by 21! for the possible ways to put in the consonants.

The summation can be evaluated by mapping this to a lattice path problem. Suppose you make a total of 26 steps of which five must be toward the North and 21 to the East. After a step to the North at least $\displaystyle d-1$ steps to the East must be made before another step to the North can be made. It is easy to see that the summation counts all possible lattice paths of this type, the summation variables indicating where a step to the North is taken.

This problem can be mapped to a problem where no restriction exists but where there are a total of $\displaystyle 26 - 4\times(d-1)$ steps. You just put in $\displaystyle d-1$ steps toward the East inbetween two steps to map the latter problem to the former problem. The latter problem is easily solved. The number of lattice paths is:
Binomial[26-4(d-1),5]. The answer is thus:



$\displaystyle \frac{(30-4d)!}{(25-4d)!} (21)!$

 

[Denis]

Amen. Go and sin no more...