Section 12.8 class notes.
I. The Counting Principle and Permutations.
Earlier in the chapter, we introduced the Counting Principle…
“If a first experiment can be performed in M distinct ways and a second experiment can
be performed in N distinct ways then the two experiments in that specific order can be
performed in M x N distinct ways”
Example: A computer password is to consist of two lower case letters followed by four digits. Determine
how many passwords are possible if…..
(These three problems are extended examples of the counting principle.)
A) Repetition of letters and digits is permitted
____ ____ ____ ____ ____ ____ =
L
L
# #
#
#
different passwords
B) Repetition of letters and digits is NOT permitted
____ ____ ____ ____ ____ ____ =
L
L
# #
#
#
different passwords
C) The first letter must be a vowel (a, e, i, o, u) and the first digit cannot be a 0, and repetition of letters
and digits is NOT permitted.
____ ____ ____ ____ ____ ____ =
L
L
# #
#
#
different passwords
Example: Bob, Sue, Larry, Sally and Fred are waiting in line to buy concert tickets.
A) In how many different ways can they stand in line?
____ ____ ____ ____ ____ =
B) If Larry insists on being in the middle, now how many ways?
____ ____
1
____ ____ =
Larry
C) If Sue insists on being first and Bob has to be last, how many ways?
1
____ ____ ____
Sue
1
=
Bob
Permutations: A permutation is any ordered arrangement of a given set of objects.
The number of permutations of “n” distinct items is n factorial , symbolized n!, where
n!  n   n 1   n  2   n  3  3   2  1
P is the number of permutations possible when r objects are selected from n objects.
n r
n
Pr 
n!
 n  r !
ON YOUR CALCULATOR!!!!!!
Example: How many different three letter “words” can be created from the letters S, T, A, R, P
This problem can be solved using the permutation formula, or by using the counting principle. I prefer the
counting principle, but both methods will work.
P 
5 3
5!
 5  3 !

______ _______ _______
Example: If a club consisting of eight members wishes to randomly select a president, vice-president and
secretary, how many different arrangements are possible? This also can be solved using the permutation
formula or the counting principle.
P 
8 3
______ _______ _______
Permutations of Duplicate Objects.
The number of distinct permutations of n objects where n1 of the objects are identical, n2 of the objects
are identical,……., nr of the objects are identical is found by the formula…
n!
n1 ! n2 ! n3 ! nr !
Example: In how many ways can the letters in the word “TALLAHASSEE” be arranged?
(there really isn’t a nice way to solve this using the counting principle and we are stuck with this formula)
Work: Of the 11 letters, three are A’s, two are L’s, two are S’s and two are E’s.
So the total number of distinct arrangements are
11!

3!2!2!2!
We can now try some problems from the text.
Section 12.8 problem 22
The daily double at most racetracks consists of selecting the winning horse in both the first and the second
race. If the first race has 7 entries and the second has 8 entries, how many daily double tickets mush you
purchase to guarantee a win?
Section 12.8 problem 26
A social security number consists of nine digits. How many different social security numbers are possible
if repetition of digits is permitted?
Section 12.8 problem 29
The operator of Sound Great Stereo store is planning a grand opening. He wishes to advertise that the store
has many different sound systems available. It stocks 8 different CD players, 10 different receivers, and 9
different sets of speakers. Assuming that a sound system will consist of one of each item, and all pieces are
compatible, how many different sound systems can they advertise?
Section 12.8 problem 33
If a club consists of 10 members, how many different arrangements of president, vice-president and
secretary are possible?
In exercises 37 – 40, an identification code is to consist of two letters followed by 4 digits. How many
different codes are possible if
38) Repetition is permitted
40) The first letter must be A, B, C or D and repetition is not permitted?
In exercises 41 – 44, a license plate is to consist of three digits followed by two uppercase letters.
Determine the number of different license plates possible if
42) Repetition of numbers and letters is not permitted
44) The first digit cannot be a zero, and repetition is not permitted
Section 12.9 Combinations
When the order of selection of the items is important to the final outcome, the problem is a
PERMUTATION problem…
Example: A president, vice-president and secretary are to be selected at random from David, Dan, Julie,
Amanda and Steve. How many ways can this be done?
______ _______ _______ =
Pres
VP
Sec
OR 5 P3 
Note: (Dan, Julie, Steve) is DIFFERENT than (Julie, Steve, Dan)
When the order of selection is NOT important to the final outcome the problem is a combination problem.
The COUNTING PRINIPLE CANNOT be used when ORDER DOES NOT MATTER
The number of combinations possible when r objects are selected from n objects is
n
n!
n

 r !r !
Cr 
Example: How many different ways can three people from Dave, Dan, Julie, Amanda and Steve be
selected to attend a meeting? Let’s try to list all of the groups, to check to make sure the formula works
5
C3 
5!

 5  3!3!
Note: (Dan, Julie Steve) is the SAME as (Julie, Steve, Dan)
Example: 10 GCC students have applied for a scholarship. 6 students will be chosen to receive this
scholarship, how many different ways can these 6 be chosen? Because the order that you were selected
doesn’t matter, and only the fact that you are selected matters this is a combination problem.
10
C6 
Example: At a Chinese restaurant, dinner for eight people consists of 3 items from column A, 4 items from
column B and 3 items from column C. If columns A, B and C have 5, 7 and 6 items respectively (from
which to choose) how many different dinner combinations are possible? This is where things get very
confusing; I am using the counting principle for this problem, along with the combination formula.
Knowing how to combine rules can be tricky.
C C C 
choose 3 items choose 4 items choose 3 items
from col A
from col B
from col C
(5 choices)
(7 choices)
(6 choices)
Let’s try some problems from the book.
Section 12.9 probems:
22) An ice cream parlor has 20 different flavors. Cynthia orders a banana split and has to select 3 different
flavors. How many different selections are possible?
24) A textbook search committee is considering 8 books for possible adoption. The committee has
decided to select three of the eight for further consideration. In how many ways can they do so?
26) Whiled visiting NYC, the Nygents want to attend 3 plays out of 10 plays they would like to see. IN
how many ways can they do so?
30) Neo Anderson wants to purchase 6 different CDs, but only has enough money to purchase 4. In how
many ways can he select 4 or 6 CDs for purchase?
34) On an English test, Tito must write an essay for 3 of 5 questions in part 1 and four of 6 questions in
part 2. How many different combinations of questions can he answer?
36) At a medical research center an experimental drug is to be given to 12 people, 6 men and 6 women. If
10 men and 9 women have volunteered to be given the drug, in how many ways can the researcher choose
the 12 people to be given the drug?
38) Michael is sent to the store to get 5 different bottles of regular soda and 3 different bottles of diet soda.
If there are 10 different types of regular soda, and 7 different types of diet soda to choose from, how many
different choices does Michael have?
POKER: The number of possible 5 card poker hands that can be dealt from a deck of 52 cards is
52 C5 
(There are 52 total cards ; 13 “types” 2, 3, 4, 5, 6, 7, 8, 9, 10, J,Q,K,A and 4 “suits” hearts, diamonds, clubs
and spades)
different hands, figure out how many of those are each of the following….
Of the
A) 1 Pair (XXABC)
C
choose the type
of pair
(pair of 3’s,
pair of J’s ?)
C
choose two
of those 4
cards
3D,3H?
C
C
C
C

choose 3
lastly, for EACH of the three
different
last cards, choose one of the
cards from possible “suits”
the remaining
12 “types”
B) 2 Pair (XXYYA)
C
C
C
C
C
choose 2 of the
13 “types” to be
your XX YY
two pairs
choose a choose a
“suit” for “suit” for
EACH card EACH card
in your first in your second
pair XX
pair YY
C) 3 of a kind
(XXXAB)
C
C
C
choose the one choose 3 of
“type” for XXX the possible
four suits for
XXX
C
D) 4 of a kind (XXXXA)
C
C
C
C
choose your 5 th card
to be one of the OTHER
11 remaining types
NOT X and NOT Y
C
choose the last
two cards A,B
to be different
than X



lastly choose a suit
for each of A and B
choose a “suit
for the 5th card
choose the “type” choose all
choose the 5th
of card for XXXX four suits
card from the
(is this needed?) remaining 12
“types”
finally, choose
a suit for your
5 th card
E) Full House (XXXYY)
C
C
C
C

choose the “type” choose 3
choose the card
of card for XXX of the 4 “suits” “type” for YY
for XXX
from the remaining
12 “types”
choose 2 of the
possible 4 suits
for YY
F) Royal Straight Flush (10,J,Q,K,A ALL of the same suit)
C 
(all we have to do in this one is choose a “suit” , the cards are already determined)
(that is to say there are only
royal straight flushes… 10 – A of hearts, diamonds, clubs and spades)
G) Straight Flush ( five consecutive cards from A,2,3,4,5,6,7,8,9,10,J,Q,K,A…all of the same suit BUT
NOT a Royal Straight Flush)
C
choose a “starting”
card for your 5 card
straight
A,2,3,4,5,6,7,8,9,10


choose 1 of
the four
suits for
the straight

subtract out
the
royal
straight flushes
that would have
been counted in the first part
H) Straight ( five consecutive cards from A,2,3,4,5,6,7,8,9,10,J,Q,K,A…NOT all of the same suit)
C
choose a “starting”
card for your 5 card
straight
A,2,3,4,5,6,7,8,9,10
C
C
C
C
now choose a “suit” for each
of the 5 cards in your
straight


subtract out the hands that
would be straight flushes (
and royal straight flushes (
)
)
I) Flush ( five cards all of the same suit BUT NOT a straight flush or a royal straight flush)
C
choose a suit
for your flush
C

choose any 5
cards of the
13 “types”

subtract out those flushes
that end up being straight flushes (
and royal straight flushes (
)
)