1.3 The refined count \(\beta _n(S)\)
For \(D \subseteq \{ 0, \dots , n-1\} \),
\[ \beta (D) = \# \{ \sigma \in \mathfrak {S}_{n+1} : D(\sigma ) = D\} . \]
This is Stanley’s \(\beta _n(S)\) from EC1 eq. (1.32), p. 30.
\(\beta (\emptyset ) = 1\) (only the identity has empty descent set) and \(\beta (\{ 0, \dots , n-1\} ) = 1\) (only the order-reversing permutation has full descent set).
\(\sum _{D \subseteq \{ 0, \dots , n-1\} } \beta (D) = (n+1)!\).
For each \(0 \leq k \leq n\),
\[ \sum _{|D| = k} \beta (D) = A_{n+1,k+1}. \]
Aggregating \(\beta \) over descent sets of fixed size recovers the Eulerian number.
\(\beta (D) = \beta (\, \rev _n \cdot D^{c}\, )\), where \(D^{c}\) denotes the set complement and \(\rev _n\) acts pointwise. In particular the two alternating descent sets have equal \(\beta \)-value.