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**Unformatted text preview: **2.3 Permutations and Combinations Permutations and combinations give us quick, algebraic methods of counting. They
are used in probability problems for two purposes: to count the number of equally likely possible results for the classical approach to probability, and to count the
number of different arrangements of the same items to give a multiplying factor. (a) Each separate arrangement of all or part of a set of items is called a permutation. The
number of permutations is the number of different arrangements in which items can
be placed. Notice that if the order of the items is changed, the arrangement is differ-
ent, so we have a different permutation. Say we have a total of n items to be arranged,
and we can choose r of those items at a time, where r s n. The number of permuta-
tions of 11 items chosen r at a time is written ”Pr. For permutations we consider both
the identity of the items and their order. Let us think for a minute about the number of choices we have at each step
along the way. If there are n distinguishable items, we have n choices for the ﬁrst
item. Having made that choice, we have (n—l) choices for the second item, then
(n — 2) choices for the third item, and so on until we come to the r th item, for
which we have (n — r + 1) choices. Then the total number of choices is given by
the product (n)(n — l)(n — 2)(n — 3)...(n — r + 1). But remember that we have a
short-hand notation for a related product, (n)(n — 1)(n — 2)(n — 3)...(3)(2)(l) = n!,
which is called n factorial or factorial n. Similarly, r! = (r)(r — 1)(r — 2)(r — 3)...
(3)(2)(l), and (n — r)! = (n — r)(n — r— l) ((n — r— 2)...(3)(2)(1). Then the total
number of choices, which is called the number Ofpennutations of n items taken
r at a time, is P _ n: _ n(n—1)(n—2)...(2)(l) '1 r (rt—r)! (n—r)(n—r—1)...(3)(2)(l)
By definition, 0! = 1. Then the number of choices of n items taken n at a time is
P” = n!. (2.6) 1’! Example 2. 13 An engineer in technical sales must Visit plants in Vancouver, Toronto, and Winnipeg. How many different sequences or orders of visiting these three plants
are possible? Answer: The number of different sequences is equal to 3P3 = 3! = 6 different
permutations. This can be verified by the following tree diagram: First Second Third Route T w vrw
V< w T vwr V w W
T< w V va T V wrv
w< V—T WVT Figure 2.15: Tree Diagram for Visits to Plants (b) The calculation of permutations is modiﬁed if some of the items cannot be
distinguished from one another. We speak of this as calculation of the
number of permutations into classes. We have already seen that if n items are
all different, the number of permutations taken n at a time is n!. However, if
some of them are indistinguishable from one another, the number of possible
permutations is reduced. If n1 items are the same, and the remaining (rt—r11) items are the same of a different class, the number of permutations can be n! .
shown to be — . The numerator, 11!, would be the number of permutations nIQ—my
of n distinguishable items taken n at a time. But :11 of these items are 1 indistinguishable, so reducing the number of permutations by a factor 5 ,
1. and another (n — n1) items are not distinguishable from one another, so reducing
l (n — n1)! '
:1 items, of which 111 are the same of one class, n2 are the same of a second class, and n3 are the same of a third class, such that n1+ n2+ n3: 1, the number
m the number of permutations by another factor If we have a total of of permutations is , . This could be extended to further classes. | |
n].n2.n3. Example 2.14 A machinist produces 22 items during a shift. Three of the 22 items are defective and
the rest are not defective. In how many different orders can the 22 items be arranged
if all the defective items are considered identical and all the nondefective items are
identical of a different class? Answer: The number of ways of arranging 3 defective items and 19 nondefective items is 22! :W : 1540. (30(19!) (3)(2)(1) (c) Combinations are similar to permutations, but with the important difference
that combinations take no account of order. Thus, AB and BA are different
permutations but the same combination of letters. Then the number of
permutations must be larger than the number of combinations, and the ratio
between them must be the number of ways the chosen items can be arranged. Say on an examination we have to do any eight questions out of ten. The
. . 10!
number of permutations of questions would be 10P8 = 3 Remember
that the number of ways in which eight items can be arranged is 8!, so the . . 1
number of combinations must be reduced by the factor —I . Then the number ' l
of combinations of 10 distinguishable items taken 8 at a time is [1% é]. In
general, the number of combinations of n items taken r at a time is
P n!
”C, = " r =— (2.7) r! (n — r) I r!
"C, gives the number of equally likely ways of choosing r items from a group of
n distinguishable items. That can be used with the classical approach to probabil- ity. ...

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- Spring '17