9.2 The Multiplication Principle, Permutations, and Combinations
Counting plays a major role in probability. In this section we shall look at special types of
counting problems and develop general formulas for solving them. We begin with an example
that will demonstrate a general counting principle.
1. The fixed-price dinner at Bubba’s Barbeque provides the following choices:
Appetizer: Soup or Salad
Entrée: Chicken, Beef Brisket, Ribs, or Sausage
Dessert: Apple Pie or Peach Cobbler
How many different meals can be served up at Bubba’s?
Choose an appetizer
2 choices
Choose an Entrée
4 choices
Choose a Dessert
2 choices
Look at the tree diagram. We see that, for each choice of appetizer there are 4 choices of entrees.
And for each of these 2(4) = 8 choices, there are 2 choices for dessert. A total of
2 · 4 · 2 16
different meals can be ordered.
Multiplication Principle of Counting
If a task consists of a sequence of choices in which there are p selections for the first choice, q
selections for the second choice, r selections for the third choice, and so on, then the task of
making these selections can be done in
· · ·…
different ways.
1. Joe has 5 shirts, 6 trousers, 3 ties, and 4 sport coats. How many different arrangements can he
wear?
Solution. By the counting principle he has 5 · 6 · 3 · 2 180 different arrangements.
2. How many different 4-letter radio station call letters can be made
a) if the first letter must be a K or W and no letter may be repeated?
b) if repeats are allowed (but the first letter is a K or W)
c) How many of the 4-letter call letters (starting with K or W) with no repeats end in R
a) By the counting principle: 1 · 25 · 24 · 23 1 · 25 · 24 · 23 27,600
b) 1 · 26 · 26 · 26 1 · 26 · 26 · 25 35,152
c) 1 · 24 · 23 · 1 1 · 24 · 23 · 1 1104
3. The United States Postal Service currently uses 5-digit zip codes in most areas. How many
zip codes are possible if there are no restrictions on the digits used? How many would be
possible if the first number could not be 0?
Solution: There are 10 digits possible for the 1st digit, 10 possible digits for the 2nd and so on.
Therefore there are 10 · 10 · 10 · 10 · 10 100,000 possible zip codes. If the first number can
not be a 0 then there are only 9 possible for the 1st and 10 possible for the remaining four
9 · 10 · 10 · 10 · 10 90,000 possible zip codes
4. For many years the state of California used 3 letters followed by 3 digits on its automobile
license plates.
a) How many different license plates are possible with this arrangement?
b) When the state ran out of new numbers, the order was reversed to 3 digits followed by 3
letters. How many new license plate numbers were then possible?
c) Several years ago, the numbers described in part b) were all used up. The state then issued
plates with1 letter followed by 3 digits and then 3 letters. How many new license plate
numbers will this provide?
Solution: a) 26 · 26 · 26 · 10 · 10 · 10 17,576,000 license plates
b) 10 · 10 · 10 · 26 · 26 · 26 17,576,000 new license plates
c) 26 · 10 · 10 · 10 · 26 · 26 · 26 456,976,000 new license plates
5. A polka band will play 6 traditional pieces and 4 original pieces at a local concert. In how
many ways can the musicians arrange the program if
a) they begin with a traditional piece?
b) they play an original piece last.
Solution: a) There are 6 traditional pieces that they can begin with leaving 9 choices for the 2nd ,
8 choices for the 3rd, and so on:
6 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 2,177,280 possible arrangements
b) There are 4 original pieces that they can end with leaving 9 choices for the 1st, 8
choices for the 2nd, 7 choices for the 3rd, and so on:
4 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 1,451,520 possible arrangements
Definition: If n is a nonnegative integer then n factorial, written as n!, is defined as
!
1
2
3
2·1
We define 0 factorial to be 1, that is, 0! 1
Evaluate the following:
7. 5!
Solution: 5! 5 · 4 · 3 · 2 · 1
120
8. 12!
Solution: 12!
·2·1
12 · 11 · 10 ·
479,001,600
A permutation of a set of values is an arrangement of those values into a particular order. As an
example the set 1,2,3 has the following six different permutations
1,2,3 , 1,3,2 , 2,1,3 , 2,3,1 , 3,1,2 ,
3,2,1
The symbol nPr represents the number of ordered arrangements of r objects chosen from n
distinct objects, where
and repetition is not allowed is given by the formula
!
nPr
!
The symbol nPr is the number of permutations of r objects chosen from n objects.
9. A baseball team has 15 players. How many 9-player batting orders are possible
!
Solution: 15P9
!
!
·
·
·
·
!
·
· · · · · · · · ·
· · · · ·
1,816,214,400
10. A student activity club at the college has 32 members. In how many different ways can the
club select a president, a vice president, a treasurer, and a secretary?
Solution: 32P4
!
!
!
!
·
·
·
·
·
·
·
·… ·
·
·… ·
32 · 31 · 30 · 29
863,040
11. A suitcase contains 6 distinct pairs of socks and 4 distinct pairs of pants. If a traveler
randomly picks 2 pairs of socks and then 3 pairs of pants, how many ways can this be done?
Solution: 6P2·4P3
!
!
!
·
!
!
!
!
·
· · · · ·
720
!
In permutations, order is important, for example, the arrangements abc, cab, bac,… are
considered different arrangements of the letters a, b, and c. In many situations, though, order is
unimportant. For example, in the card game of poker, the order in which the cards are received
does not matter; it is the combination of the cards that matters.
A combination is an arrangement, without regard to order, of r objects selected from n distinct
objects without repetition.
12. List all combinations of the 4 objects a,b,c,d taken 2 at a time.
Solution ab,ac,ad,bc,bd, and cd. There are 6 combinations of the 4 objects taken 2 at a time.
The number of combinations of n elements taken r at a time, where
, is
!
nCr
! !
13. A group of 4 students is to be selected from a group of 10 students to take part in a class in
cell biology.
a) In how many ways can this be done?
b) In how many ways can the group that will not take part be chosen?
Solution: a) 10C4
b) 10C6
!
!
· · · · · ·…· ·
· · ·
! !
! !
!
· ·…· · · · · ·
· · ·
! !
! !
!
210
210
14. From a group of 15 smokers and 21 nonsmokers, a researcher wants to randomly select 7
smokers and 6 nonsmokers for a study. In how many ways can the study group be selected?
Solution: 15C7·21C6
!
! !
·
!
!
! !
! !
·
!
! !
349,188,840
15. Seven cards are drawn from an ordinary deck. In how many ways is it possible to draw (a)
only 4’s (b) only 6’s, 7’s, and 8’s (c) no 6’s, 7’s, and 8’s (d) exactly 2 kings or jacks; (e) 2 spades
and 5 hearts?
Solution a) 0
b) 12C7 = 792
c) 40C7 = 18,643,560
d) 8C2·44C5 = 30,408,224
e) 13C2·13C5 = 100,386
16. A class has 10 male students and 12 female students. How many ways can the class select a
committee of four people to petition the teacher not to make the final exam cumulative if the
committee has to have 2 males and 2 females?
Solution 10C2·12C2 2970