What you should learn:
Goal
1
Use Combinations to count the number of ways an event can happen.
Goal
2
Use the Binomial Theorem to expand a binomial that is raised to a power.
10.2 Combinatins and Binomial Theorem

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
12.2 Combinatins and Binomial Theorem

A Combination is a selection of r objects from a group of n objects where order is not important
12.2 Combinatins and Binomial Theorem

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 n
C r and is: n
C r
 n !
( n
 r )!
r !
12.2 Combinatins and Binomial Theorem

Ex. 1) the number of combinations of 2 objects taken from a group of 5 objects is
5
C
2

2
5 !
( 5

2 )!
2 !

5 * 4 * 3 * 2 * 1

10
3 * 2 * 1 * 2 * 1
12.2 Combinatins and Binomial Theorem

Finding Combinations
• In a standard deck of 52 cards there are 4 suits with 13 of each suit.
Ex. 2a) 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
C
5

52 !
( 52

5 )!
5 !

52 * 51 * 50 * 49 * 48 * 47 !
47 !* 5 !
= 2,598,960

Ex. 2b.) 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
C
1
*
13
C
5

4 !
3 !* 1 !
*
13 !
8 !* 5 !

4 * 3 !
*
3 !* 1 !
13 * 12 * 11 * 10 * 9 * 8 !
8 !* 5 !

5148
12.2 Combinatins and Binomial Theorem

3a.) How many 7 card hands are
52
C
7
 possible?
52 !

133 , 784 , 560
45 !* 7 !
• 3b.) How many of these hands have all 7 cards the same suit?
4
C
1
*
13
C
7

6864
12.2 Combinatins and Binomial Theorem

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 +.
12.2 Combinatins and Binomial Theorem

Deciding to ADD or MULTIPLY
Ex. 4a.) 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
C
2
*
4
C
1

6 !
4 !
2 !
*
4 !
3 !
1 !

15 * 4

60
12.2 Combinatins and Binomial Theorem

•
10-2B Binomial Theorem

Ex. 4b) 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
C
0

10
C
1

10
C
2

10
C
3

1

10

45

120

176
12.2 Combinatins and Binomial Theorem

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.
12.2 Combinatins and Binomial Theorem

Subtracting instead of adding:
Ex. 5)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:
• Or……
12
C
3

12
C
4

12
C
5

...

12
C
12

For each play you can attend you can go or not go with 2 options.
• So, like section 10.1 it would be
2*2*2*2*2*2*2*2*2*2*2*2 =2 12
• And you will not attend 0, or 1, or 2.
• So:
2
12 
(
12
C
0

12
C
1

12
C
2
)

4096

( 1

12

66 )

4017





0
C
0
1
C
0 1
C
1

2
C
0 2
C
1 2
C
2
3
C
0 3
C
1 3
C
2 3
C
3
4
C
0 4
C
1 4
C
2 4
C
3 4
C
4

Etc…
12.2 Combinatins and Binomial Theorem

Pascal's Triangle!
Which adds 2 #’s together of each row for the next row
• 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 = a 2 + 2ab + b 2 (1 2 1)
• (a+b) 3 = a 3 (b 0 )+3a 2 b 1 +3a 1 b 2 +b 3 (a 0 )
(1 3 3 1)
• (a+b) 4 = a 4 +4a 3 b+6a 2 b 2 +4ab 3 +b 4
(1 4 6 4 1)
• In general…
12.2 Combinatins and Binomial Theorem

(a+b) n (n is a positive integer)=
• n
C
0 a n b 0 + n
C
1 a n-1 b 1 + n
C n a 0 b n n
C
2 a n-2 b 2 + …+
• = r n 

0 n
C r a n
 r b r
12.2 Combinatins and Binomial Theorem

Ex. 6 (a+3) 5

Ex. 6 (a+3) 5 =
•
5
C
0 a 5 3 0 +
5
C
1 a 4 3 1 +
5
C
2 a 3 3 2 +
5
C
3 a 2 3
3 +
5
C
4 a 1 3 4 +
5
C
5 a 0 3 5 =
• 1a 5 + 15a 4 + 90a 3 + 270a 2 + 405a
+ 243
12.2 Combinatins and Binomial Theorem

12.2 Combinatins and Binomial Theorem