12 2 combinations and binomial thm revised

12.2 Combinations

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 timeThe number of combinations of r objects taken from a group of n distinct objects is denoted by nCr and is:

!)!(

!

rrn

nCrn

For instance, the number of combinations of 2 objects taken from a group of 5 objects is

101*2*1*2*3

1*2*3*4*5

!2)!25(

!525

C

Finding CombinationsIn 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

!5!*47

!47*48*49*50*51*52

!5)!552(

!52552

C

= 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:

5148!5!*8

!8*9*10*11*12*13*

!1!*3

!3*4

!5!*8

!13*

!1!*3

!4* 51314 CC

How many 7 card hands are possible?

How many of these hands have all 7 cards the same suit?

560,784,133!7!*45

!52752 C

6864* 71314 CC

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?

604*15!1!3

!4*

!2!4

!6* 1426 CC

Suppose you can afford at most 3 ingredientsHow many different types can you order?

You can order an omelet w/ 0, or 1, or 2, or 3 items and there are 10 items to choose from.

17612045101310210110010 CCCC

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:

Or……1212512412312 ... CCCC

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:

4017)66121(4096)(2 21211201212 CCC