Figuring out how to create sets of

number or groups of elements, and

expand binomials

*

Combination: a selection of r

objects from a group of n objects

where order is not important.

!

( )! !n r

nC

n r r

How is this formula different from the one

for permutations, nPr?

How is this formula the same?

Memorize this formula!!!

Example 1: How many different ways can

you choose a group of 7 books from a shelf

containing 32 books if order isn’t important?

32C7= ?

Remember to use parentheses

around the denominator or use

the nCr choice on the

calculator.

Ex 2: There are 12 comedies, 8 action

movies, 7 dramas, 5 suspense and 9

family movies.

a) In how many different ways can you

choose exactly 2 comedies and 3

family movies?

SOL: 12C2· 9C3

(if AND is involved,

you multiply the

combinations,

if OR is involved,

you add combinations)

There are 12 comedies, 8 action

movies, 7 dramas, 5 suspense and 9

family movies.

b) You can afford at most 2 movies. How

many movie combinations could you go

home with?

You could rent 0, 1, or 2

movies. There are 41

movies total.

So 41C0 + 41C1 + 41C2

Ex 3: There are 11 articles in a magazine.

You have time to read at least 2 articles.

How many different combinations of articles

could you choose?

One way: 11C2 + 11C3 +11C4 + . . . + 11C11

(choose 2 articles or 3 articles or 4

articles . . . or 11 articles)

Another way: Since you can choose to

read or not read each article, there

are 2 choices.

211 – (11C0 + 11C1 )

(you didn’t read just 1 or 0 articles.)

Now you try one: There are 10 games

played in a soccer season. How many

combinations of games could you attend if

you chose at least 3 of them?

10C3 + 10C4 +10C5 + . . . + 10C10

or

210 – (10C0 + 10C1 + 10C2 ) = 968(2 choices each time - go or not go)

Pascal’s Triangle: Pascal was a mathematician

who noticed a pattern that ended up to be

useful!0C0

1C0 1C1

2C0 2C1 2C2

3C0 3C1 3C2 3C3

n=0

n=1

n=2

n=3

(a+b)0= 1

(a+b)1= 1a + 1b

(a+b)2= 1a2 + 2ab + 1b2

(a+b)3= 1a3 + 3a2b + 3ab2 +1b3

In the last line, notice the pattern of the

coefficients - they come from Pascal’s triangle, from

the row that has 3 as the second number. Also

notice the pattern of the powers of a and b – a goes

down in power, b goes up in power.

n=0

n=1

n=2

n=3

Binomial Theorem:

For any positive integer n,

0 0 1 1 2 2

0 1 2( ) ...

...

n n n n n r r

n n n n r

n n n

n n

a b C a b C a b C a b C a b

C a b

You can see the pattern of the coefficients that

come from Pascal’s triangle, from the row that

has n as the second number. You can also see the

pattern of the powers of a and b –

a goes down in power, b goes up in power.

0 0 1 1 2 2

0 1 2( ) ... ...n n n n n r r n n n

n n n n r n na b C a b C a b C a b C a b C a b

Ex 4: Expand (2x + 1)4

4 0 0 4 1 1 4 2 2 4 3 3 4 4 4

4 0 4 1 4 2 4 3 4 4(2 ) (1) (2 ) (1) (2 ) (1) (2 ) (1) (2 ) (1)C x C x C x C x C x

4 0 3 1 2 2 1 3 0 41(2 ) (1) 4(2 ) (1) 6(2 ) (1) 4(2 ) (1) 1(2 ) (1)x x x x x

4 3 2 11(16 )(1) 4(8 )(1) 6(4 )(1) 4(2 ) (1) 1(1)(1)x x x x

4 3 216 32 24 8 1x x x x

I think it is far faster to generate Pascal’s

triangle and pick the row for coefficients

based on the power than to put nCr into my

calculator each time, but you can choose to do

it either way.

Ex 5: Expand (x – 2y)3

3 3 0 2 1 1 2 0 3( 2 ) 1( ) ( 2 ) 3( ) ( 2 ) 3( ) ( 2 ) 1( ) ( 2 )x y x y x y x y x y

Row from Pascal’s that has 3 as second number:

1 3 3 1

3 2 2 31( )(1) 3( )( 2 ) 3( )(4 ) 1(1)( 8 )x x y x y y

3 2 2 36 12 8x x y xy y

And last but not least,

Ex 6: Find the coefficient of the term containing

x4 in (2x – 7)9

We start with where r=term#-1.

Powers in the polynomial would be

9,8,7,6,5,4,3,2,1, so x4 would be in the 6th term.

r = 6 – 1, so r = 5

(9-r has to equal 4)

9

9 (2 ) ( 7)r r

rC x

4 5126(2 ) ( 7)x

9 5 5

9 5(2 ) ( 7)C x

4126(16 )( 16807)x

The coefficient is -33,882,912

You can try it: Find coefficient of x4 in the

expansion of (3x – 1)11

This is #24 on 10.2 B assignment!

11 7 7

11 7(3 ) ( 1)C x