Modern Engineering Thermodynamics

18.7 Introduction to Mathematical Probability 745
The last mathematical concepts needed in our study of probability are those of permutations and combinations.
A specific ordered arrangement of N distinguishable objects is called a permutation. The total number of ways of
making different ordered arrangements of the N objects taken R at a time using no object more than once is
given by
P N R = N!
ðN − RÞ!
(18.35a)
where N! = NðN − 1ÞðN − 2ÞðN − 3Þ … ð3Þð2Þð1Þ is N factorial. Notethatwedefine0!= 1. Thus, the total
number of permutations of N distinct objects taken N at a time is
P N R
= N!/0! = N!
(18.35b)
However, if the objects are allowed to be repeated within an arrangement, then the total number of arrangements
becomes
P N N = NR
(18.35c)
Forexample,thetotalnumberofpermutations(arrangements)ofthetendigits0,1,2,3,4,5,6,7,8,and9
taken three at a time, using no digit more than once, is given by Eq. (18.35a) with N = 10 and R = 3as
ðP3 10 Þ =
10!
using each
ð10 − 3Þ! = 10! =
10 × 9 × 8 × 7!
= 720 arrangements
7! 7!
digit only once
But, if we allow each digit to be used more than once, then Eq. (18.35c) tells us that the number of permutations
increases to
ðP3 10 Þ = 10 3 = 1000 arrangements
using each
digit more than once
However, the total number of permutations (arrangements) of just the three digits 1, 2, and 3, using none of
these digits more than once is given by Eq. (18.35b) as
ðP3 3 Þ = 3! = 3 × 2 × 1 = 6 arrangements
using each
digit only once
(they are 1, 2, 3; 1, 3, 2; 3, 1, 2; 2, 3, 1; 3, 2, 1; and 2, 1, 3), but if we allow these digits to be repeated, the
Eq. (18.35c) shows that the number of permutations increases dramatically to
ðP3 3 Þ = 3 3 = 27 arrangements
using each
digit more than once
We define combinations as the ways of choosing a sample of R objects from a group of N objects without regard
to order within the sample (e.g., the groupings AB and BA are different permutations of A and B, buttheyare
the same combination of A and B). The total number of combinations of N unique objects taken R at a time
without using any object more than once is given by
C N R = PN R
R!
=
N!
ðN − RÞ!R!
(18.36a)
But, if the objects are allowed to be repeated within the sample, then the number of combinations becomes
C N R = PN R
R!
=
ðN + R − 1Þ!
ðN − 1Þ!R!
(18.36b)
For example, we can determine the number of five-card hands that can be dealt out of a standard deck of 52
distinct cards (Table 18.6). Since the order in which the cards are received is not important, the number of combinations
of 52 cards taken 5 at a time is given by Eq. (18.36a) as
ðC 52
5 Þ =
52!
=
52 × 51 × 50 × 49 × 48 × 47!
= 2,598,960 different hands
using each
ð52 − 5Þ!5! 47! × 5!
card only once