A practical difficulty that sometimes arises in computing the probability of an event is counting the numbers of basic outcomes in the sample space and the event of interest. For some problems the use of permutations or combinations can be helpful.
1. Number of Orderings
We begin with the problem of ordering. Suppose that we have some number x of objects that are to be placed in order. Each object may be used only once. How many different sequences are possible? We can view this problem as a requirement to place one of the objects in each of x boxes arranged in a row.
3.2 Probability and Its Postulates 103 Beginning with the left box in Figure 3.5, there are x different ways to fill it. Once an object is put in that box, there are 1x - 12 objects remaining, and so 1x - 12 ways to fill the second box. That is, for each of the x ways to place an object in the first box, there are 1x - 12 possible ways to fill the second box, so the first two boxes can be filled in a total of x1x - 12 ways. Given that the first two boxes are filled, there are now 1x - 22 ways of filling the third box, so the first three boxes can be filled in a total of x1x - 121x - 22 ways. When we arrive at the last box, there is only one object left to put in it. Finally, we arrive at the number of possible orderings.
(x – 1) (x – 2)
. . .
2x 1
Figure 3.5 The Orderings of x Objects
Number of Possible Orderings
The total number of possible ways of arranging x objects in order is given by x1x - 121x - 22 g 122112 = x!
where x! is read “x factorial.”
2.Permutations
Suppose that now we have a number n of objects with which the x ordered boxes could be filled (with n 7 x). Each object may be used only once. The number of possible orderings is called the number of permutations of x objects chosen from n and is denoted by the symbol Pnx.
We can argue precisely as before, except that there will be n ways to fill the first box, 1n - 12 ways to fill the second box, and so on, until we come to the final box. At this point there will be 1n - x + 12 objects left, each of which could be placed in that box, as illustrated in Figure 3.6.
Figure 3.6 The Permutations of x Objects
Chosen From n Objects
n (n – 1) (n – 2)
. . .
(n – x + 2) (n – x + 1)(n – x) objects left over
Permutations
The total number of permutations of x objects chosen from n, Pnx, is the num-ber of possible arrangements when x objects are to be selected from a total of n and arranged in order.
Pnx = n1n - 121n - 22 g1n - x + 12 Multiplying and dividing the right hand side by
1n - x21n - x - 12 g122112 = 1n - x2!
gives
Pnx = n1n - 121n - 22 g1n - x + 121n - x21n - x - 12 g122112 1n - x21n - x - 12 g122112
= n!
1n - x2!
104 Chapter 3 Elements of Chance: Probability Methods
Example 3.6 Five Letters (Permutations)
Suppose that two letters are to be selected from A, B, C, D, and E and arranged in order.
How many permutations are possible?
Solution The number of permutations, with n = 5 and x = 2, is as follows:
P52 = 5!
3! = 20 These are
AB AC AD AE BC
BA CA DA EA CB
BD BE CD CE DE
DB EB DC EC ED
3.Combinations
Finally, suppose that we are interested in the number of different ways that x objects can be selected from n (where no object may be chosen more than once) but order is not impor-tant. Notice in Example 3.6 that the entries in the second and fourth rows are just rear-rangements of those directly above them and may, therefore, be ignored. Thus, there are only 10 possibilities for selecting two objects from a group of 5 if order is not important.
The number of possible selections is called the number of combinations and is denoted by Cnx; here x objects are to be chosen from n. To find this number, note first that the number of possible permutations is Pnx. However, many of these will be rearrangements of the same x objects and, therefore, are irrelevant. In fact, since x objects can be ordered in x!
ways, we are concerned with only a proportion 1>x! of the permutations. This leads us to a previously stated outcome—namely, Equation 3.5.
Number of Combinations
The number of combinations, Cnx, of x objects chosen from n is the number of possible selections that can be made. This number is
Cnx = Pnx x!
or, simply,
Cnx = n!
x!1n - x2! (3.5)
In some applications the notation an
xb = Cnx = n!
x!1n - x2!
is used.
We illustrate the combination equation, Equation 3.5, by noting that in Example 3.5 the number of combinations of the 5 computers taken 2 at a time is the number of ele-ments in the sample space:
C52 = 5!
2!15 - 22! = 5
#
4#
3#
2#
1 2#
113#
2#
12 = 103.2 Probability and Its Postulates 105
Example 3.7 Probability of Employee Selection (Combinations)
A personnel officer has 8 candidates to fill 4 similar positions. 5 candidates are men, and 3 are women. If, in fact, every combination of candidates is equally likely to be chosen, what is the probability that no women will be hired?
Solution First, the total number of possible combinations of 4 candidates chosen from 8 is as follows:
C84 = 8!
4!4! = 70
Now, in order for no women to be hired, it follows that the 4 successful candidates must come from the available 5 men. The number of such combinations is as follows:
C54 = 5!
4!1! = 5
Therefore, if at the outset each of the 70 possible combinations was equally likely to be chosen, the probability that one of the 5 all-male combinations would be selected is 5>70 = 1>14.
Example 3.8 Computer Selection Revised (Classical Probability)
Suppose that Karlyn’s store now contains 10 Hewlett-Packard computers, 5 Dell puters, and 5 Sony computers. Susan enters the store and wants to purchase 3 com-puters. The computers are selected purely by chance from the shelf. Now what is the probability that she selects 2 Hewlett-Packard computers and 1 Dell?
Solution The classical definition of probability will be used. But in this example the combinations formula will be used to determine the number of outcomes in the sample space and the number of outcomes that satisfy the condition A: [2 Hewlett-Packard and 1 Dell].
The total number of outcomes in the sample space is as follows:
N = C203 = 20!
3!120 - 32! = 1,140
The number of ways that we can select 2 Hewlett-Packard computers from the 10 available is computed by the following:
C102 = 10!
2!110 - 22! = 45
Similarly, the number of ways that we can select 1 Dell computer from the 5 available is 5 and, therefore, the number of outcomes that satisfy event A is as follows:
NA = C102 * C51 = 45 * 5 = 225
Finally, the probability of A = [2 Hewlett-Packard and 1 Dell] is as follows:
PA = NA
N = C102 * C51
C203 = 45 * 5
1,140 = 0.197
106 Chapter 3 Elements of Chance: Probability Methods