combinations, binomial theorem

1. Figuring out how to create sets of
number or groups of elements, and
expand binomials
*

2. Combination: a selection of r
objects from a group of n objects
where order is not important.
!
( )! !
n r
n
C
n r r

 
How is this formula different from the one
for permutations, nPr?
How is this formula the same?
Memorize this formula!!!

3. 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.

4. 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)

5. 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

6. 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.)

7. 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)

8. 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

9. 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

10. 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.

11. 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 4
1(2 ) (1) 4(2 ) (1) 6(2 ) (1) 4(2 ) (1) 1(2 ) (1)x x x x x    
4 3 2 1
1(16 )(1) 4(8 )(1) 6(4 )(1) 4(2 ) (1) 1(1)(1)x x x x    
4 3 2
16 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.

12. 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 3
1( )(1) 3( )( 2 ) 3( )(4 ) 1(1)( 8 )x x y x y y     
3 2 2 3
6 12 8x x y xy y   

13. 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 5
126(2 ) ( 7)x 
9 5 5
9 5(2 ) ( 7)C x 

4
126(16 )( 16807)x 
The coefficient is -33,882,912

14. 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 
