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How can I show for example AB+BC+AC simply. It is adding up the permutations of n numbers. Another example would be ABC+ABD+ACD+BCD.

Sorry I'll try to make it clear with an example ( which is sort of the problem is know the terminology to use)

If you have three people a, b, c working and they each build a house in A, B and C days respectively then the equation for the number of days for them all working together to build the one house is ABC/(AB+BC+AC). If you have n number of people then it is ABC.....( all n days are multiplied) divided by the combinations added up.
There is "shorthand" i.e. a symbol, for the top but I am asking if there is a shorthand for the bottom of this. This type of equation is used in may applications that imply 'working together' such as resistance, springs....

Boof
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  • What do you mean by "addition of permutations." Such an operation is not usually defined. There is composition of permutations, but that is a different concept. – JMoravitz May 25 '16 at 05:08
  • It's not clear what you mean by "adding the permutations of $n$ numbers". Do you mean $12+23+13=48$? This doesn't seem to be well-defined because $AB$ and $BA$ are different values. – svsring May 25 '16 at 07:21

1 Answers1

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Given a set of variables, $x_1,x_2,\dots,x_n$ the Elementary Symmetric Polynomial $e_k(x_1,x_2,\dots,x_n)$ is the summation of all distinct products of $k$ distinct variables from $x_1,\dots,x_n$.

For example, $e_2(a,b,c) = ab+ac+bc$

$e_3(a,b,c,d)=abc+abd+acd+bcd$

$e_2(a,b,c,d) = ab+ac+ad+bc+bd+cd$

It is worth mentioning yet again that these are not permutations.

Old answer before clarification was added

A permutation is defined as a bijective function from a set to itself.

For example, if our set is $\{dog,cat\}$ we have the following two permutations:

$f~:~\{dog,cat\}\to\{dog,cat\}$ given by $f(dog)=dog$ and $f(cat)=cat$

$g~:~\{dog,cat\}\to\{dog,cat\}$ given by $g(dog)=cat$ and $g(cat)=dog$

Given an order on the elements of the set (say, $dog<cat$)These can be expressed more compactly as a sequence: $dog,cat$ and $cat,dog$.

$f$ and $g$ are the permutations here.

$f+g$ doesn't make sense. After all, what is $dog+cat$? We only cared that we had a set to begin with. We do not care in the slightest whether or not operations are defined on that set. Addition of permutations is generally not defined (at least not in the usual sense).

If your question was how to count the number of permutations of an $n$-element set, the answer is $n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n$.

If your question was how to count the number of length $k$ strings made from elements of an $n$-element set with no repetition, the answer is $\frac{n!}{(n-k)!}$. For example, with the set $\{A,B,C\}$ and $k=2$ we have the following: $AB,AC,BC,BA,CA,CB$

If your question was how to count the number of subsets of size $k$ of an $n$-element set, the answer is $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. For example, the two-element subsets of the set $\{A,B,C\}$ are $\{A,B\},\{A,C\},\{B,C\}$ and the three-element subsets of the set $\{A,B,C,D\}$ are $\{A,B,C\},\{A,B,D\},\{A,C,D\},\{B,C,D\}$

Given that the "permutations" you were "adding" together in your question match those I write in the last paragraph, my best guess to your question was that you meant to ask how to count the number of subsets of size $k$ from an $n$-element set. Until you confirm or edit and properly explain what it is you are wanting to ask however, noone can be sure.

JMoravitz
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  • thanks for you help - see edited version of question for an example. – Boof May 25 '16 at 23:53
  • @Boof now that you have properly explained your question, I have added relevant information in my recent edit. – JMoravitz May 26 '16 at 00:22
  • Excellent. Thanks. An now I have a name for them. – Boof May 26 '16 at 05:18
  • So a general equation for the problem with n workers ( springs resistance etc) is en/e(n-1) where n and n-1 are subscripts (the formatting of which I haven't got time to figure out). Nice elegant. – Boof May 26 '16 at 05:27
  • @Boof Formatting tips can be found here. Also, $e_n$ by itself is meaningless in this context. $e_n(x_1,x_2,\dots,x_n)$ and $e_{n-1}(x_1,x_2,\dots,x_n)$. This notation in my experience is commonly found in abstract algebra in the study of galois theory and representation theory. Be warned that many people will not be familiar with the notation as it is not frequently used in other contexts. – JMoravitz May 26 '16 at 17:30