6.1 Use Combinations
and Binomial Theorem
378
What is a combination?
How is it different from a permutation?
What is the formula for nCr?
In the last section we learned
counting problems where
order was important
For other counting problems where
order is NOT important like cards, (the
order you’re dealt is not important,
after you get them, reordering them
doesn’t change your hand)
 These unordered groupings are called
Combinations

A Combination is a selection of
“r” objects from a group of “n”
objects where order is not
important
Combination of n objects
taken r at a time

The number of combinations of r
objects taken from a group of n distinct
objects is denoted by nCr and is:
n!
C

n r
(n  r )! r!

For instance, the number of
combinations of 2 objects taken from a
group of 5 objects is
2
5!
5 * 4 * 3 * 2 *1

 10
5 C2 
(5  2)!2! 3 * 2 *1* 2 *1
Finding Combinations
In a standard deck of 52 cards there
are 4 suits with 13 of each suit.
 If the order isn’t important how many
different 5-card hands are possible?
 The number of ways to draw 5 cards
from 52 is
52!
52 * 51* 50 * 49 * 48 * 47!

52 C5 
(52  5)!5!
47!*5!
= 2,598,960

In how many of these hands
are all 5 cards the same suit?
You need to choose 1 of the 4 suits and
then 5 of the 13 cards in the suit.
 The number of possible hands are:

4!
13! 4 * 3! 13 *12 *11*10 * 9 * 8!
*

*
 5148
4 C1 *13 C5 
3!*1! 8!*5! 3!*1!
8!*5!
How many 7 card hands are
possible?
52!
 133,784,560
52 C7 
45!*7!

How many of these hands have all 7
cards the same suit?
C
*
C

6864
4 1 13 7
Theater
William Shakespeare wrote 38 plays that can be
divided into three genres. Of the 38 plays, 18 are
comedies, 10 are histories, and 10 are tragedies.
a. How many different sets of exactly 2
comedies and 1 tragedy can you read?
a.
You can choose 2 of the 18 comedies and 1 of the 10
tragedies. So, the number of possible sets of plays
is:
18!
10!
C
C
=
18 2 10 1
16!
2!
9! 1!
18 17 16!
16! 2 1
= 153 10
= 1530
=
10
9!
9!
1
b.
How many different sets of at most 3 plays
can you read?
b.
You can read 0, 1, 2, or 3 plays. Because there are
38 plays that can be chosen, the number of
possible sets of plays is:
38C0
+ 38C1 + 38C2 +38C3 = 1 + 38 + 703 + 8436
= 9178
 When
finding the number of
ways both an event A and an
event B can occur, you multiply.
 When finding the number of
ways that an event A OR B can
occur, you +.
Deciding to + or *
A restaurant serves omelets. They offer
6 vegetarian ingredients and 4 meat
ingredients.
 You want exactly 2 veg. ingredients and
1 meat. How many kinds of omelets
can you order?

6! 4!
*
 15 * 4  60
6 C 2 *4 C1 
4!2! 3!1!
Suppose you can afford at
most 3 ingredients
How many different types can you
order?
 You can order an omelet with 0, or 1,
or 2, or 3 items and there are 10 items
to choose from.

10
C0 10 C1 10 C2 10 C3  1  10  45  120  176
Counting
problems that
involve ‘at least’ or ‘at most’
sometimes are easier to
solve by subtracting
possibilities you don’t want
from the total number of
possibilities.
Subtracting instead of adding:
A theatre is having 12 plays. You want
to attend at least 3. How many
combinations of plays can you attend?
 You want to attend 3 or 4 or 5 or … or
12.
 From this section you would solve the
problem using:


12
Or……
C3 12 C4 12 C5  ... 12 C12
 For
each play you can attend
you can go or not go.
 So, like section 12.1 it would be
2*2*2*2*2*2*2*2*2*2*2*2
=212
 And you will not attend 0, or 1,
or 2.
 So:
212  (12 C0 12 C1 12 C2 )  4096  (1  12  66)  4017
Basketball
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?
SOLUTION
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: 12C3 + 12C4 + 12C5 +………..+ 12C12
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:
212 – (12C0 + 12C1 + 12C2=) 4096 – (1 + 12 + 66)
= 4017
6.1 Assignment Day 1
 Page
382, 3-18 all, 38-40 all
6.1 Use Combinations and
Binomial Theorem Day 2
What is Pascal’s Triangle?
What is the Binomial Theorem?
How is Pascal’s Triangle connected to
the binomial theorem?
Pascal’s Triangle

If you arrange the values
of nCr in a triangular
pattern in which each row
corresponds to a value N,
you get what is called
Pascal’s triangle. Pascal’s
triangle is named after the
French mathematician
Blaise Pascal (1623-1662)
The Binomial Theorem
 0C0
 1C0 1C1
 2C0 2C1 2C2
 3C0 3C1 3C2 3C3
 4C0 4C1 4C2 4C3 4C4
 Etc…
Pascal's Triangle!

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
 1 5 10 10 5 1
 Etc…
 This describes the coefficients in the
expansion of the binomial (a+b)n

 (a+b)2
= a2 + 2ab + b2 (1 2 1)
 (a+b)3 =
a3(b0)+3a2b1+3a1b2+b3(a0)
(1 3 3 1)
 (a+b)4 = a4+4a3b+6a2b2+4ab3+b4
(1 4 6 4 1)
 In general…
(a+b)n (n is a positive integer)=
 nC0anb0 + nC1an-1b1 + nC2an-2b2

=
n
C
a
n r
r 0
nr
b
r
+ …+ nCna0bn
(a+3)5 =
5C0a530+5C1a431+5C2a332+5C3a233
1
4
0
5
+5C4a 3 +5C5a 3 =
1a5 + 15a4 + 90a3 +
405a + 243
270a2 +
(a+3)5 =
 1a530+5a431+10a332+10a233+5a134
+1a035
 1a5 + 15a4 + 90a3 + 270a2 + 405a
+ 243
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
School Clubs
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.
SOLUTION
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)
n = 6 (6th row)
1
1
5
6
6C0 6C1
10
15
6C2
10
20
6C3
6C4
5
1
15
6C5
6
6C6
ANSWER
The value of 6C2 is the third number in the 6th
row of Pascal’s triangle, as shown above.
Therefore, 6C2 = 15. There are 15 combinations
of representatives for the convention.
Use the binomial theorem to write the binomial
expansion.
(x2 + y)3 = 3C0(x2)3y0 + 3C1(x2)2y1 + 3C2(x2)1y2 + 3C3(x2)0y3
= (1)(x6)(1) + (3)(x4)(y) + (3)(x2)(y2) + (1)(1)(y3)
= x6 + 3x4y + 3x2y2 + y3
Find the coefficient of x4 in the expansion of (3x + 2)10.
SOLUTION
From the binomial theorem, you know the following:
(3x + 2)10 =10C0(3x)10(2)0 + 10C1(3x)9(2)1 + . . . + 10C10(3x)0(2)10
Each term in the expansion has the form
10 – r (2) r. The term containing x4 occurs
C
(3x)
10 r
when r = 6:
4(2)6
4)(64) = 1,088,640x4
C
(3x)
=
(210)(81x
10 6
ANSWER
The coefficient of x4 is 1,088,640.
Find the coefficient of x5 in the expansion of (x – 3)7.
SOLUTION
From the binomial theorem, you know the following:
(x – 3)7= 7C0(x)7(–3)0 + 7C1(x)6(–3)1 + . . . + 7C7(x)0(–3)7
Each term in the expansion has the form 7Cr(x)7 – r
(–3)r. The term containing x5 occurs when r = 2:
7Cr(x)
7 – r (–3)r
ANSWER
= (21)(x5)(9) = 189x5
The coefficient of x5 is 189.
Find the coefficient of x3 in the expansion of (2x + 5)8.
SOLUTION
From the binomial theorem, you know the following
(2x + 5)8 = 8C0(2x)8(5)0 + 8C1(2x)7(5)1 + . . . + 8C8(2x)0(5)8
Each term in the expansion has the form 8Cr(2x)8 – r (5)r.
The term containing x3 occurs when r = 5:
3(5)5 = (56)(8x3)(3125)
3
C
(2x)
=
1,400,000x
8 5
ANSWER
The coefficient of x3 is 1,400,000.
What is Pascal’s Triangle?
An arrangement of the values of nCr in a triangular
pattern.
 What is the Binomial Theorem?
A formula that gives the coefficients in the raising of
(a+b) to a power.
 How is Pascal’s Triangle connected to the binomial
theorem?
Pascal’s Triangle gives the same coefficients as the
Binomial Theorem.

Assignment 12.2
Page
382
20-30even, 32-34 all,48-50
all