Chapter 10
Ways to Count- Tree Diagram

A sporting goods store offers 3 types of
snowboards (all-mountain, freestyle, and
carving) and 2 types of boots (soft and
hybrid). How many choices does the
store offer for snowboarding equipment?
A sporting goods store offers 3 types of snowboards (all-mountain,
freestyle, and carving) and 2 types of boots (soft and hybrid). How
many choices does the store offer for snowboarding equipment?
Soft
All-mountain
board, soft
boots
Hybrid
all-mountain
board, hybrid
boots
Soft
Freestyle
board, soft
boots
Hybrid
Freestyle
board, hybrid
boots
Soft
Carving board,
soft boots
Hard
Carving board,
hard boots
All-mountain
Freestyle
Carving
Ways to count- Tree Diagram

The tree has 6 branches. So, there are 6
possible choices!
Use a Tree Diagram

The snowboard shop also offers 3
different types of bicycles (mountain,
racing, and BMX) and 3 different wheel
sizes (20”, 22”, and 24”). How many
bicycle choices does the store offer?
ANSWER

9 Bicycles
Fundamental Counting Principle

Snowboarding example There are 3 options for boards and 2 options
for boots.
 So the number of possibilities is 32=6
This is called the “Fundamental Counting
Principle”
Fundamental Counting Principle

Two events If one event can occur in m ways and
another event can occur in n ways, then the
number of ways that both events can occur
is mn.

Three or more events The fundamental counting principle extends
to 3 or more events. The product of
possibilities for each event will give the
possibility of ALL the events occurring.
Use the Fundamental Counting
Principle

You are framing a picture. The frames
are available in 12 different styles. Each
style is available in 55 different colors.
You also want a blue mat board, which
is available in 11 different shades of
blue. How many different ways can you
frame the picture?
Use the Fundamental Counting
Principle

Multiply
 # of frame styles (12)
 # of frame colors (55)
 # of mat boards (11)
12 x 55 x 11 = 7260
There are 7260 different ways to frame the
picture!
Use the Counting Principle with
or without Repetition

The standard configuration for a Texas
license plate is 1 letter followed by 2
digits followed by 3 letters
Use the Counting Principle with
Repetition

How many different license plates are
possible if letters and digits can be
repeated?
Use the Counting Principle with
Repetition
How many different license plates are
possible if letters and digits can be
repeated?
 Multiply







# of letters possible (26)
# of digits possible (10)
# of digits possible (10)
# of letters possible (26)
# of letters possible (26)
# of letters possible (26)
Use the Counting Principle with
Repetition

26 x 10 x 10 x 26 x 26 x 26 =
45,697,600
Use the Counting Principle
without Repetition

How many different license plates are
possible if letters and digits cannot be
repeated?
Use the Counting Principle
without Repetition


How many different license plates are possible
if letters and digits cannot be repeated?
Multiply






# of letters possible (26)
# of digits possible (10)
# of digits possible w/o the previous digit used (9)
# of letters possible w/o the first letter used (25)
# of letters possible w/o the first 2 letters used (24)
# of letters possible w/o the first 3 letters used (23)
Use the Counting Principle
without Repetition

26 x 10 x 9 x 25 x 24 x 23 = 32,292,000
More Counting Principles with
and without Repetition
How would the answers change for the
standard configuration of a New York
license plate, which is 3 letters followed
by 4 numbers?
 How many possibilities with repetition?
 How many possibilities without
repetition?

Answer
With Repetition- 175,760,000
possibilities
 Without Repetition- 78,624,000
possibilities

Permutations

Question- How many ways can you
rearrange the word
COW
COW
CWO
OCW
OWC
WOC
WCO
Permutations
The answer is 6 times.
 The ordering of a certain number of n
objects is called a PERMUTATION.

Permutations
Back to the question… “How many ways
can you rearrange the word COW?”
 You can use the fundamental counting
principle to find the number of
permutations of C, O, and W.
 First letter- 3 choices
 Second letter- 2 choices
 Third letter- 1 choice

Permutations

Therefore, we find the number of
permutations is 3 x 2 x 1 = 6
Permutations

How many permutations for the word:
SEA
ROCK
MATH
Factorials!
3 x 2 x 1 can also be written as 3!
 ! is the symbol used for factorial
 3! is read “three factorial”

Factorials!

In general, n! is defined where n is a
positive integer as follows:
n! = n  (n-1)  (n-2)  …  3  2  1
Factorials!

5! = 5  4  3  2  1 = 120

12! = 12  11  10  9  8  7  6  5  4 
3  2  1 = 479,001,600
Permutations using Factorials

Ten teams are competing in the final
round of the Olympic four-person
bobsledding competition. In how many
different ways can the bobsledding
teams finish the competition? (Assume
no ties)
Permutations with Factorials
There are 10! ways that teams can
finish.
 10! = 10  9  8  7  6  5  4  3  2  1
= 3,628,800

Permutations with Factorials

How many different ways can 3 of the
bobsledding teams finish first, second,
and third to win the gold, silver, and
bronze medals?
Permutations with Factorials
Any of the 10 teams can finish first, then
any of the remaining 9 teams can finish
second, and finally any of the 8
remaining teams can finish third.
 So number of ways teams can win
medals is

10  9  8 = 720
Permutations with Factorials
How would the answers change if there
were 12 bobsledding teams competing
in the final round of he competition?
 A- How many was can the teams finish?
 B- How many ways to win medals?

Answer
A- 12! = 479,001,600
 B- 12  11  10 = 1320

Permutations of n Objects Taken
r at a Time
The answer to the second part of the
bobsledding questions is called the number of
permutations of 10 objects taken 3 at a time.
 Denoted as 10P3
 10P3 = 10  9  8 = 10 ·9 ·8· 7·6·5· 4·3·2 ·1
7·6 ·5· 4·3·2·1
10!
10!
=
=

7!
(10 - 3)!
Permutations with Repetition

The general equation for the number of
permutations of n objects taken r at a
time is:
n!
n Pr =
(n - r)!
Use the Permutation
Equation

You a burning a demo CD for your band.
Your band has 12 songs stored on your
computer. However, you want to put only
4 songs on the demo CD. In how many
orders can you burn 4 of the 12 songs
onto the CD?
Use the Permutation
Equation
12!
12! 479, 001, 600
=
=
=11,880
12 P4 =
(12 - 4)! 8!
40,320

Try:

5P3
 4P1
 8P5
Practice Answers
 5P 3 =
60
 4P 1 =
4
 8P 5 =
6720
Permutations with Repetition

How many ways can you write the word
MOO
Note: Each “O” is a distinct letter.
Permutations with Repetition
MOO
MOO
OMO
OMO
OOM
OOM
There are 6 permutations of the letters M,
O, and O.
However, if the two occurrences of O are
considered interchangeable, then there
are only three distinguishable
permutations: MOO OMO OOM
Permutations with Repetition
Each of these permutations corresponds to
two of the original six permutations because
there are 2! , or 2, permutations of O and O.
 So the number of permutations of M, O, and
O can be written as 3! = 6 = 3

2!
2
Permutations with Repetition

To generalize the situation we just
explored we can say that the number of
distinguishable permutations of n
objects where one object is repeated s1
times, another is repeated s2 times, and
so on, is:
n!
s1 !· s2 !·...· sk !
Permutations with Repetition

How many ways can you rearrange the
letters of the following words:
MATHEMATICS
 ALGEBRA
 MISSISSIPPI
 TALLAHASSEE

Permutations with Repetition

MATHEMATICS
11!
39, 916,800
=
= 4, 989, 600
2!·2!·2!
2 ·2 ·2

ALGEBRA
7! 5, 040
=
= 2, 520
2!
2
Permutations with Repetition

MISSISSIPPI
11!
39, 916,800
=
= 34, 650
4!· 4!·2! 24·24·2

TALLAHASSEE
11!
39, 916,800
=
= 831, 600
3!·2!·2!·2! 6·2·2 ·2
Permutations vs Combinations
With permutations, we learned that
order is important for some counting
problems.
 For other counting problems, order is
not important.
 For example, if you purchase a package
of trading cards, the order of the cards
inside the package is not important.

Permutations vs. Combinations
COMBINATION- a selection of r objects
from a group of n objects where order is
NOT important.
 Think of a combination pizza; you have
various kinds of toppings (pepperoni,
olives, green peppers, etc.) and you can
put them on in several different orders,
but in the end, order doesn’t matter. Its
just one delicious pizza!

Combination Equation

The number of combinations of r objects
taken from a group of n distinct objects
is denoted by nCr and is given by this
formula:
n!
n Cr =
(n - r)!·r!
 nC r
is read “n choose r”
Find Combinations
A standard deck of 52 playing cards has
4 suits with 13 different cards in each
suit.
 SUIT- hearts, clubs, spades, and
diamonds

Find Combinations

Each suit has the following cards called
“kinds”:
 Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen,
King
 There is one of each of these kinds in each
suit

Face Cards- Cards that have faces on
them (Jack, Queen, King)
Find Combinations
If the order in which the cards are dealt
is not important, how many different 5card hands are possible?
 Remember…
n!

n

Cr =
(n - r)!·r!
In this example, n= 52 cards and r = 5
cards
Find Combinations

Therefore,
52 C5 =
52!
52!
52 ·51·50· 49· 48· 47!
=
=
= 2, 598, 960
(52 - 5)!·5! 47!·5!
47!·5!
Find Combinations

In how many 5-card hands are all 5
cards of the same color?
Find Combinations

For all 5 cards to be the same color, you
need to choose 1 of the 2 colors and
then 5 of the 26 cards in that color. So,
the number of possible hands is:
2!
26!
2 26 ·25·24·23·22 ·21!
·
=
·
=131, 560
2 C 1· 26 C5 =
1!·1! 21!·5! 1·1
21!·5!
Multiple Events
When finding the number of ways both
an event A and an event B can occur,
you need to multiply.
 When finding the number of ways that
event A or event B can occur, you add
instead.

Add or Multiply Combinations

William Shakespeare wrote 38 plays
that can be divided into 3 genres. Of the
38 plays, 18 are comedies, 10 are
histories, and 10 are tragedies.
 How many different sets of exactly 2
comedies and 1 tragedy can you read?
 How many different sets of at most 3 plays
can you read?
Add or Multiply Combination
How many different sets of exactly 2
comedies and 1 tragedy can you read?
 You can choose 2 of the 18 comedies
and 1 of the 10 tragedies. So, the
number of possible sets of plays is:

18 C 2 ·10 C1 =
18!
10! 18·17·16! 10·9!
·
=
·
=153·10 =1, 530
16!·2! 9!·1! 16!·2·1
9!·1
Add or Multiply Combinations
How many different sets of at most 3
plays can you read?
 You could read 0, 1, 2, or 3 plays.
Because there are 38 plays that can be
chosen, the number of possible sets of
plays is:

38
C0 + 38 C1 + 38 C2 + 38 C3 =1+38+ 703+8, 436 = 9,178
Practice
How many different sets of exactly 3
tragedies and 2 histories can you read?
 8C 3

 10C6
 7C 2
Practice Answers

5400 sets
 8C 3 =
56
 10C6 =
 7C 2 =
210
21
Subtracting Possibilities
Counting problems that involve phrases
like “at most” or “at least” are sometimes
easier to solve by subtracting
possibilities you do not want from the
total number of possibilities.
 The portion NOT included in the number
of possibilities is also called the
COMPLIMENT.

Subtracting Possibilities

During the school year, the girl’s
basketball team is scheduled to play 12
home games. You want to attend at
least 3 of the games. How many
different combinations of games can
you attend?
Subtracting Possibilities

Of the 12 home games, you want to
attend 3 games, or 4 games, or 5
games, and so on. So, the number of
combinations of games you can attend
is:
C + C + C +... + C
12
3
12
4
12
5
12
12
Subtracting Possibilities

Instead of adding these combinations,
use the following reasoning. For each of
the 12 games, you can choose to attend
or not attend the game, so there are 212
total combinations. If you attend at least
3 games, you do not attend only a total
of 0, 1, or 2 games. So, the number of
ways you can attend at least 3 games is:
2 - ( 12 C0 + 12 C1 + 12 C2 ) = 4, 096 - (1+12 + 66) = 4, 017
12
Pascal’s Triangle

If you arrange the values of nCr in a
triangular pattern in which each row
corresponds to a value of n, you get
what is called Pascal’s triangle, named
after the French mathematician Blaise
Pascal.
Pascal’s Triangle
Pascal’s triangle is show below with its entries
represented by combinations and with its entries
represented by numbers The first and last
numbers in each row are 1. Every number other
than 1 is the sum of the closest two numbers in
the row directly above it.
Pascal’s Triangle as
numbers
Pascal’s Triangle as
combinations
Use Pascal’s Triangle

The 6 members of a Model UN club
must choose 2 representatives to attend
a state convention. Use Pascal’s triangle
to find the number of combinations of 2
members that can be chosen as
representatives.
Use Pascal’s Triangle
Because you need to find 6C2, write the
6th row of Pascal’s triangle by adding
numbers from the previous row.
 n = 5 (5th row) 1 5 10 10 5 1


n = 6 (6th row) 1 6 15 20 15 6 1

The value of 6C2 is the 3rd number in the 6th row
of Pascal’s triangle, as show above. Therefore,
6C2 = 15. There are 15 combinations of
representatives for the convention.
Binomial Expansions

There is an important relationship
between powers of binomials and
combinations. The numbers in Pascal’s
triangle can be used to find coefficients
in binomial expansions.
Binomial Expansions
For example, the coefficients in the
expansion (a + b)4 are the numbers of
combinations in the row of Pascal’s
triangle for n = 4:
(a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4

4C 0
4C1
4C2
4C3
4C4
This result is generalized in the binomial
theorem.
Binomial Theorem

For any positive integer n, the binomial
expansion of (a + b)n is:
(a + b)n = nC0anb0 + nC1an-1b1 + nC2an-2b2 +…+ nCna0bn

Notice that each term in the expansion
of (a + b)n has the form nCran-rbr where r
is an integer from 0 to n.
Expand a power of a binomial
sum

Use the binomial theorem to write the
binomial expansion for:
(x2 + y)3
Expand a power of a binomial
sum
(x 2 + y)3 = 3 C0 (x 2 )3 y0 + 3 C1(x 2 )2 y1 + 3 C2 (x 2 )1 y2 + 3 C3 (x 2 )0 y3
= (1)(x 6 )(1)+ (3)(x 4 )(y)+ (3)(x 2 )(y2 )+ (1)(1)(y3 )
= x + 3x y + 3x y + y
6
4
2 2
3
Expand a power of a binomial
difference

Use the binomial theorem to write the
binomial expansion of:
(a-2b)4
Expand a power of a binomial
difference
(a - 2b)4 = [a + (-2b)]4
= 4 C0 a4 (-2b)0 + 4 C1a3 (-2b)1 + 4 C2 a2 (-2b)2 + 4 C3a1 (-2b)3 + 4 C4 a0 (-2b)4
= (1)(a4 )(1)+(4)(a3 )(-2b)+(6)(a2 )(4b2 )+(4)(a)(-8b3 )+(1)(1)(16b4 )
= a -8a b + 24ab - 32ab +16b
4
3
2
3
4
Powers of Binomial Differences

To expand a power of a binomial
difference, you can rewrite the binomial
as a sum. The resulting expansion will
have terms whose signs alternate
between + and -.
Use the binomial theorem to
write the binomial expansion

(x + 3)5

(a + 2b)4

(2p – q)4

(5 – 2y)3

(x + 3)5
= 5 C0 a5 (3)0 + 5 C1a4 (3)1 + 5 C2 a3 (3)2 + 5 C3a2 (3)3 + 5 C4 a1 (3)4 + 5 C5a0 (3)5
= (1)a 5 (1) + (5)a 4 (3) + (10)a3 (9) + (10)a 2 (27) + (5)a(81) + (1)(1)(243)
= a 5 +15a 4 + 90a3 + 270a 2 + 405a + 243

(a + 2b)4
= 4 C0 a4 (2b)0 + 4 C1a3 (2b)1 + 4 C2 a2 (2b)2 + 4 C3a1 (2b)3 + 4 C4 a0 (2b)4
= (1)(a4 )(1)+(4)(a3 )(2b)+(6)(a2 )(4b2 )+(4)(a)(8b3 )+(1)(1)(16b4 )
= a +8a b + 24ab + 32ab +16b
4
3
2
3
4
Find a coefficient in an expansion

Find the coefficient of x4 in the
expansion of (3x + 2)10
Find a coefficient in an expansion

From the binomial theorem, you know
the following:
(3x + 2) = 10 C0 (3x) (2) + 10 C1 (3x) (2) +... + 10 C10 (3x) (2)
10
10
0
9
1
Each term in the expansion has the form
10-r
r
4
C
(3x)
(2)
The
term
containing
x
10 r
occurs when r = 6:

4
6
4
4
C
(3x)
(2)
=
(210)(81x
)(64)
=1,
088,
640x
10 6
0
10
Find the coefficient in an
expansion
So the coefficient of x4 is 1,088,640
 Find the coefficient of:

 x5 in the expansion of (x – 3)7
 x3 in the expansion of (2x + 5)8
Practice Answers

x5 in the expansion of (x – 3)7
 189

x3 in the expansion of (2x + 5)8
 1,400,000