By Richard P. Stanley

This e-book, the 1st of a two-volume easy creation to enumerative combinatorics, concentrates at the conception and alertness of producing services, a primary software in enumerative combinatorics. Richard Stanley covers these elements of enumerative combinatorics with the best purposes to different parts of arithmetic. The 4 chapters are dedicated to an available creation to enumeration, sieve methods--including the main of Inclusion-Exclusion, partly ordered units, and rational producing services. a number of workouts, just about all with options, increase the textual content and supply access into many components no longer lined without delay. Graduate scholars and study mathematicians who desire to practice combinatorics to their paintings will locate this an authoritative reference.

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**Example text**

N , where γk is a permutation of {w1 , . . , wk }. First, let γ1 = w1 . Assume that γk has been deﬁned for some 1 ≤ k < n. If the last letter of γk (which turns out to be wk ) is greater (respectively, smaller) than wk+1 , then split γk after each letter greater (respectively, smaller) than wk+1 . These splits divide γk into compartments. Cyclically shift each compartment of γk one unit to the right, and place wk+1 at the end. Let γk+1 be the word thus obtained. Set ϕ(w) = γn . 7 Example. Before analyzing the map ϕ, let us ﬁrst give an example.

Though irrelevant here, it is interesting to note that this sum is just the (n − k)th elementary symmetric function of 1, 2, . . 29) counts the number of pairs (S, f ), where S ∈ [n−1] n−k and f : S → [n − 1] satisﬁes f (i) ≤ i. Thus, we seek a bijection φ : → Sn,k between the set of all such pairs (S, f ), and the set Sn,k of w ∈ Sn with k cycles. Given (S, f ) ∈ where S = {a1 , . . , an−k }< ⊆ [n − 1], deﬁne T = {j ∈ [n] : n − j ∈ S}. Let the elements of [n] − T be b1 > b2 > · · · > bn−k .

Write F (t; x) = F (t1 , t2 , . . 26). 4 Example. Let e6 (n) be the number of permutations w ∈ Sn satisfying w 6 = 1. A permutation w satisﬁes w 6 = 1 if and only if all its cycles have length 1,2,3 or 6. Hence, e6 (n) = n! Zn (ti = 1 if i|6, ti = 0 otherwise). There follows e6 (n) n≥0 xn = F (ti = 1 if i|6, ti = 0 otherwise) n! = exp x + x2 x3 x6 + + . 31). 5 Example. Let Ek (n) denote the expected number of k-cycles in a permutation w ∈ Sn . It is understood that the expectation is taken with respect to the uniform distribution on Sn , so Ek (n) = 1 n!