Number of permutations

[cybergeo3]

I'll try to explain this as simply as possible. I am wondering how to calculate the number of permutations of two variables across 26 elements with the distribution of variables required to be even. That's confusing, so let me illustrate. Let's say I have the letters A-Z. 26 elements. I assign each letter the number 1 or 2. For the first permutation, Let's assign A-M the number "1" and N-Z the number "2". There are exactly 13 "1"s and 13 "2"s. For a second permutation, let's say A-L are "1", M-Y are "2" and Z is "1". This is a different permutation, but there are still exactly 13 of each.

A-Z all being assigned 1 is an invalid permutation in this example, as there must be exactly 13 of each variable.

Hopefully that explains the permutations I am trying to calculate. So my question is, how would I calculate the number of permutations that could exist in this example?

Thanks!

 

[pka]

I'll try to explain this as simply as possible. I am wondering how to calculate the number of permutations of two variables across 26 elements with the distribution of variables required to be even. That's confusing, so let me illustrate. Let's say I have the letters A-Z. 26 elements. I assign each letter the number 1 or 2. For the first permutation, Let's assign A-M the number "1" and N-Z the number "2". There are exactly 13 "1"s and 13 "2"s. For a second permutation, let's say A-L are "1", M-Y are "2" and Z is "1". This is a different permutation, but there are still exactly 13 of each.
A-Z all being assigned 1 is an invalid permutation in this example, as there must be exactly 13 of each variable.
Hopefully that explains the permutations I am trying to calculate. So my question is, how would I calculate the number of permutations that could exist in this example?

Frankly I do not follow you description.
Is this the question: "How many ways are there to label twenty-six letters using 1 or 2, so that there are thirteen labeled 1"?
If that is the question, then \(\displaystyle \binom{26}{13}=10400600\).

If not, try to clarify the question.

 

[Yogi]

Sometimes when working with larger numbers it helps to do a smaller problem. If there are 4 elements the possibilities are:

1122
1212
1221
2112
2121
2211
So we get 6. We are taking four elements and choosing two of them to be "1".
order doesnt matter so...
\(\displaystyle \binom{n}{k}\) = n!/(k!*(n-k)!)
\(\displaystyle \binom{4}{2}\)= 4!/(2!*2!)=6

 

[cybergeo3]

Yes, I believe both of those are exactly the route I was looking for. Thank you! (sorry for what may have been a naive question, it's been a few years since I was in school, and I couldn't remember how to do it accurately. Greatly appreciate it!