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

29.9 Permutations and combinations 949
Solution
The various possibilitiesare listed.
Primary route
A
B
A
C
B
C
Secondary route
B
A
C
A
C
B
There are six ways in which the choices can be made. Alternatively we could argue
as follows. Suppose the primary route is chosen first. There are three choices: any
oneofA,BorC.Thesecondaryrouteisthenchosenfromthetworemainingroutes,
givingtwopossiblechoices.Togetherthereare3×2 = 6waysofchoosingaprimary
route and a secondary route.
InEngineeringapplication29.2thechoiceABisdistinctfromthechoiceBA,thatisthe
order is important. Choosing two routes from three and arranging them in order is an
example of apermutation. More generally:
Apermutationofndistinctobjectstakenr atatimeisanarrangementofr ofthen
objects.
Informingpermutations,theorderoftheobjectsisimportant.Ifthreelettersarechosen
fromthealphabetthepermutationXYZisdistinctfromthepermutationZXY.Wepose
the question ‘How many permutations are there of n objects taken r at a time?’ The
following example will helptoestablish a formulafor the number ofpermutations.
Example29.12 Calculate the number of permutations thereare of
(a) four distinctobjects taken two atatime
(b) five distinctobjects taken threeatatime
(c) seven distinctobjects taken four atatime
Solution (a) Listing all possible permutations is not feasible when the numbers involved are
large. In choosing the first object, four choices are possible. In choosing the second
object, three choices are possible. There are thus 4 × 3 = 12 permutations of
four objects taken two atatime.Note that12 may be writtenas
12=4×3= 4!
2! = 4!
(4 −2)!
(b) There are five objects available for the first choice, four for the second choice and
threeforthethirdchoice.Hencethereare5×4×3 = 60permutations.Againnote
that
60 = 5!
2! = 5!
(5 −3)!