Day 14 Class Notes:
I. The Counting Principle and Permutations.
On Day 12 (Tree Diagrams) 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…..
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
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?
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?
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!
II. 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 number of combinations possible when r objects are selected from n objects is
n
Cr 
n!
 n  r  !r !
Example: How many different ways can three people from David, Dan, Julie, Amanda and Steve be
selected to attend a meeting?
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?
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?
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)
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?
B) 2 Pair (XXYYA)
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”
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
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
D) 4 of a kind (XXXXA)
C
C
C
C

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)
choose a “suit
for the 5th card
(all we have to do in this one is choose a “suit” , the cards are already determined)
C 
(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
C
choose a “starting”
card for your 5 card
straight
A,2,3,4,5,6,7,8,9,10
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 (
)
)