By Nicholas Loehr

This textbook, geared toward starting graduate scholars, is the 1st to survey the topic emphasizing the position of bijections. ... the ultimate bankruptcy incorporates a potpourri of pleasant effects ... The exposition is cautious and planned, and leaves no stone unturned ... a welcome boost to the literature and a truly great book.-David Callan, Mathematical reports, 2012dA rule i've got came across to be actual is that any bookRead more...

summary: This textbook, aimed toward starting graduate scholars, is the 1st to survey the topic emphasizing the function of bijections. ... the ultimate bankruptcy features a potpourri of pleasant effects ... The exposition is cautious and planned, and leaves no stone unturned ... a great addition to the literature and a really great book.-David Callan, Mathematical studies, 2012dA rule i've got chanced on to be actual is that any e-book claiming to be appropriate for rookies and but resulting in the frontiers of unsolved learn difficulties does neither good. This booklet is the exception to that rule. ... i discovered this e-book attractive

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**Extra resources for Bijective Combinatorics**

**Example text**

47. Example. We now illustrate the constructions in each of the two preceding proofs. For the first proof, suppose we are counting R(a3 b1 c4 ). The alphabet A in the proof consists of the eight distinct letters A = {a(1) , a(2) , a(3) , b(1) , c(1) , c(2) , c(3) , c(4) }. Let us build a specific permutation of A using the second counting method. First, choose an element of R(a3 b1 c4 ), say v = baccaacc. Second, choose a labeling of the a’s with superscripts, say ba(3) cca(1) a(2) cc. Third, choose a labeling of the b’s, say b(1) a(3) cca(1) a(2) cc.

Flush hands. To build a typical flush H, first choose the one-element set S(H) in any of 41 = 4 ways. Then choose the five-element set V (H) in any of 13 5 ways. H is now completely determined since all cards in H have the same suit. The total is therefore 4 · 13 = 5148. The sample hand above was constructed by choosing S(H) = {♥}, 5 then V (H) = {3, 8, 10, J, Q}. In the table entry for flushes, we subtract the number of straight flushes (namely 40, as shown below) so that the entries in the table will be pairwise disjoint subsets of X.

This is not a coincidence. Indeed, we can formally define a word w1 w2 · · · wk over an alphabet A as the function w : {1, 2, . . , k} → A defined by w(i) = wi . The number of such words (functions) is |A|k . The word w1 w2 · · · wk is a k-permutation of A iff the wi ’s are all distinct iff w is an injective function. The word w1 w2 · · · wk is a permutation of A iff w is a bijective function. Finally, note that w is surjective iff every letter in the alphabet A occurs among the letters w1 , .