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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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950 Chapter 29 Statistics and probability distributions

(c) Therearesevenobjectsavailableforthefirstchoice,sixforthesecond,fiveforthe

third and four for the fourth. The number of permutations is 7 ×6×5×4 = 840.

7!

This may bewritten as

(7 −4)! .

Theexample illustratesthe following generalrule.

Thenumberofpermutationsofndistinctobjectstakenratatime,writtenP(n,r),is

P(n,r) =

n!

(n−r)!

Example29.13 Findthe numberofpermutations of

(a) 10 distinctobjects takensixatatime

(b) 15 distinctobjects takentwo atatime

(c) sixdistinctobjects takensixatatime

Solution (a) P(10,6) = 10!

(10 −6)! = 10!

4! = 151200

15!

(b) P(15,2) =

(15 −2)! = 15!

13! = 210

6!

(c) P(6,6) =

(6 −6)! = 6! = 720 (Note that0! = 1)

0!

P(6,6) issimplythe number of ways of arranging all sixof the objects.

Note that

P(n,n) =n!

This isthe numberofways ofarrangingngiven objects.

29.9.2 Combinations

Closelyrelatedto,but nevertheless distinctfrom, permutations arecombinations.

Acombination isaselection ofrdistinctobjectsfromnobjects.

Inmakingaselectiontheorderisunimportant.Forexample,giventhelettersA,BandC,

ABandBAarethesamecombinationbutdifferentpermutations.Aswithpermutations

wedevelopanexpressionforthenumberofcombinationsofnobjectstakenratatime.

Examples29.14and 29.15helpwith thisdevelopment.

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