By Koh Khee Meng, Tay Eng Guan

Publication Description

Publication Date: January 25, 2013

This ebook in its moment version is an invaluable, appealing creation to uncomplicated counting strategies for higher secondary to undergraduate scholars, in addition to academics. more youthful scholars and lay those who have fun with arithmetic, let alone avid puzzle solvers, also will locate the ebook fascinating. a few of the difficulties and functions listed below are stable for build up talent in counting. also they are beneficial for honing uncomplicated talents and strategies quite often challenge fixing. a few of the difficulties keep away from regimen and the diligent reader will frequently detect a couple of manner of fixing a specific challenge, that's certainly a big information in challenge fixing. The ebook therefore is helping to offer scholars an early begin to studying problem-solving heuristics and pondering skills.

New chapters initially from a supplementary ebook were further during this variation to considerably raise the insurance of counting strategies. the hot chapters comprise the primary of Inclusion and Exclusion, the Pigeonhole precept, Recurrence kinfolk, the Stirling Numbers and the Catalan Numbers. a few new difficulties have additionally been additional to this edition.

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**Additional resources for Counting (2nd Edition)**

**Sample text**

Thus, by (MP), the number of ways to arrange any r elements from N n in a row is given by n(n - l)(n - 2 ) . . (n - r + 1). For convenience, let us call an arrangement of any r elements from N n in a row, an r-permutation of N n , and denote by P™ the number of r-permutations of N n . Thus, we have P r " = n(n - l)(n - 2 ) . . 2. 1) for P™ looks a bit long. We shall make it more concise by introducing the following useful notation. Given a positive integer n, define n! to be the product of the n consecutive integers n, n — 1 , .

4 • 3 • 2 • 1 = 24. " is read "n factorial". By convention, we define 0! = 1. Using the "factorial" notation, we now have P r n = n(n - 1 ) . . (n - r + 1) n{n — 1 ) . . (n — r + l)(n — r)(n — r — 1 ) . . 2 • 1 _ n\ (n — r)(n — r — 1 ) . . 3) When n = 4 and r = 3, we obtain 4! r 3 — (4 - 3)! P4 - 4! 1! 4• 3• 2• 1 = 4 • 3 • 2 = 24, 1 which agrees with what we found before. 3) is valid when 0 < r < n. Consider two extreme cases: when r = 0 and r = n respectively. 3), pn __ n! = ^ = 1 (n - 0)!

1. ,n n choices n-\ choices 1 n-(r-2) n-0-l) choices choices n—2 choices ... 1 rth Counting 20 We wish to choose r elements from {1, 2 , . . ,n} to fill the r spaces, where the ordering of elements matters. There are n choices for the 1st space. After fixing one in the 1st space, there are n — 1 choices remaining for the 2nd space. After fixing one in the 2nd space, there are n — 2 choices left for the 3rd space, and so on. After fixing one in the (r — l)th space, there are n — (r — 1) choices left for the rth space.