Modern Engineering Thermodynamics

746 CHAPTER 18: Introduction to Statistical Thermodynamics
andsincethereare13possiblefour-of-a-kind hands in the deck with each hand having any of the remaining
48 cards as the fifth card in the hand, the probability of getting a five-card hand with four-of-a-kind is
P four-of-a-kind hand =
13 × 48
2,598,960 = 0:00024
or only 0.024% of the time. However, suppose each card is returned to the deck before the next card is drawn
so that a card can be drawn more than once (i.e., repeated), then the number of combinations of five-card
hands is given by Eq. (18.36b) as
ðC 52
5 Þ ð52 + 5 − 1Þ!
= =
56 × 55 × 54 × 53 × 52 × 51!
= 3,819,816 different hands
using each card ð52 − 1Þ!5!
51! × 5!
more than once
Suppose that not all objects in the group of N are different from each other. The number of permutations of
N objects, R 1 of one kind, R 2 of a second kind, … , R k of a kth kind is given by
P N R 1 ,R 2 ,…,R k
=
N!
R 1 !R 2 !R 3 ! … R k !
(18.37)
since the objects within the k groups are no longer unique, the total number of combinations and permutations
are equal, or
C N R 1 ,R 2 ,…,R k
= P N R 1 ,R 2 ,…,R k
EXAMPLE 18.7
Suppose you have to form a team of five students from a group of ten available students.
a. How many different five-person groups of officers could you form from the ten students if each student filled one of the
positions of president, vice president, secretary, treasurer, and events coordinator (i.e., the group is ordered), without
using any student more than once?
b. How many ordered officer groups could you form if you allowed students to be in more than one group?
c. How many officer groups could you form if the students were not assigned a position (i.e., the groups were not
ordered) but a student could not be in more than one group?
d. How many teams could you form if the students were not assigned a position (i.e., the groups were not ordered) and
students could be in more than one group?
e. If there are four men and six women in the group, how many different unordered ten-person groups could be formed?
Solution
a. Equation (18.35a) gives the number of ordered groups of R = 5 things chosen from a group of N = 10 as
P5
10
10!
=
using each ð10 − 5Þ! = 10! =
10 × 9 × 8 × 7 × 6 × 5!
5!
5!
student only once
= 10 × 9 × 8 × 7 × 6 = 30,240 groups
b. Equation (18.35c) gives the result when the students are allowed to be in more than one group:
ðP3 10 Þ = 10 5 = 100,000 groups
using each student
more than once
c. If the students are not assigned a position within the group, but they are allowed to belong to only one group, then
Eq. (18.36a) gives the possible number of groups as
ðC 10
5 Þ 10! 10 × 9 × 8 × 7 × 6 × 5!
=
= = 252 groups
using each
ð10 − 5Þ!5! 5! × 5!
student only once
d. If the students are not assigned a position within the group, but they are allowed to belong to more than one group,
then Eq. (18.36b) gives the possible number of groups as
ðC 10
5 Þ ð10 + 5 − 1Þ!
= =
14 × 13 × 12 × 11 × 10 × 9!
= 2002 groups
using each student ð10 − 1Þ!5!
9! × 5!
more than once