MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am interested in the problem of counting the number of linear extensions of series-parallel structures. The wikipedia article at pointed me to the following formulae:

L( P ; Q ) = L(P)L(Q)


L( P || Q ) = ( (|P| + |Q|)! / (|P|! . |Q|!) ) . L(P) . L(Q)

However, these do not cover the cases where the duplicates are omitted.

For example, for the structure

x || x

the number of linear extensions should be 1, when duplication is omitted, however the formula above would deliver 2.

Does anyone know how the factoring of the duplicates can be integrated into the formula above?

share|cite|improve this question

Normally linear extensions are counted for labelled posets. You are asking about the unlabelled case. Simply take the number of labelled linear extensions and divide by the order of the automorphism group $G$ of the poset, since in the action of $G$ on labelled linear extensions all orbits have size $|G|$. It is easy to compute $|G|$ as a product of factorials arising from parallel connections of isomorphic posets.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.