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

29.9 Permutations and combinations 951
Example29.14 There are three routes, A, B and C, joining two computers. In how many ways can two
routes be chosen from A,Band C?
Solution The possible combinations (selections) can be listedas
AB,BC, AC
thatis,therearethreepossiblewaysofmakingtheselection.Wecanalsouseourknowledge
of permutations to calculate the number of combinations. There areP(3,2) ways
of arranging the threeroutes taken two atatime.
P(3,2) =
3!
(3 −2)! = 6
EachcombinationoftworoutescanbearrangedinP(2,2)ways;thatis,eachcombinationgivesrisetotwopermutations.Forexample,thecombinationABcouldbearranged
as AB or BA, giving two permutations. Thus, the number of combinations is half the
number of permutations. There are 6/2 = 3 combinations.
Example29.15 Calculate the numberofcombinationsof
(a) sixdistinctobjectstaken fouratatime
(b) 10 distinctobjects takensixatatime
Solution (a) Consider one combination of four objects. These four objects can be arranged in
4! ways; that is, each combination gives rise to 4! permutations. The number of
permutations ofsixobjects takenfouratatime is
P(6,4) = 6!
2!
Hence the number of combinations is
P(6,4)
4!
= 6!
2!4! = 15
There are15 combinations of sixobjects taken four atatime.
(b) Eachcombination,comprisingsixobjects,givesriseto6!permutations.Thenumber
of permutations of 10 objects taken sixatatime is
P(10,6) = 10!
4!
Hence the number of combinations = P(10,6)
6!
= 10!
4!6! = 210.
( n
We write to denote the number of combinations ofnobjects takenrat a time. A
r)
( n
formulafor isnow developed.
r)