1. 2.3
Counting Partitions: Combinations

2. Combination
 Combination is just like permutation – you are
counting the number of ways to pick from a set
without repetition of elements.
 The difference is that, for combination, order
does not matter.
 Slot method can only be used when order
matters; therefore, you cannot use slots for a
combination problem.

3. Formula

n – number of things you are choosing from
r – number of things you are choosing
 Since slots cannot be used, this formula is your
only tool in solving combination problems

4.  Let’s do an example combination problem. You
order Mother Bear’s Pizza with a friend late at night.
There is a special on a 3-topping pizza so you
decide to go with that. There are 8 toppings to
choose from. How many different pizzas can
possibly be made?
 Simply apply the formula here.
 n = 8 (8 toppings to choose from)
 r = 3 (3 toppings being chosen to put on the pizza):
 Which comes to 56.

5. Quiz 2.3 #1
 How many ways can a committee of 3 be formed
from 10 club members?
A. 120
B. 240
C. 720

6. Quiz 2.3 #1
 How many ways can a committee of 3 be formed
from 10 club members?
A. 120
B. 240
C. 720
 Answer: A

7. Distinguishing between
Permutation and Combination
 Permutation – order matters
Combination – order doesn’t matter
 Let’s think of it another way – if the “slots” are
distinguishable between each other, then order
matters (P). In other words, if you rearrange the
same elements, it becomes a different set.
 Ex: A 3-digit number from {1, 2, 3, 4, 5}. This is a
permutation, since digit 1 and digit 2 are
“distinguishable.” If you switch them, it becomes a
different number, i.e. 123 is different than 213.

8.  On the other hand, if the slots are
“indistinguishable,” then order doesn’t matter (C). In
other words, if you rearrange the same elements, it
is still the same thing.
 Ex: 3 toppings from 10 on a pizza. You cannot
distinguish between topping 1 and topping 2;
Pepperoni, Sausage and Ham is the same thing as
Ham, Pepperoni and Sausage.
 One good way to identify between permutation and
combination is to go ahead and draw and label the
slots, and see if the slots are “distinguishable,” and
if switching two elements from two slots affects the
outcome.
 Remember, neither (P) or (C) can have repetitions. If
there are repetitions, use neither of these two

10.  Quiz 2.3 #2: Identify whether each
problem is permutation, combination, or neither:

11.  Quiz 2.3 #2: Identify whether each
problem is permutation, combination, or neither:
 1. Number of ways to form a committee of president,
VP, and treasurer from 10 students.

12.  Quiz 2.3 #2: Identify whether each
problem is permutation, combination, or neither:
 1. Number of ways to form a committee of president,
VP, and treasurer from 10 students.
 Permutation

13.  Quiz 2.3 #2: Identify whether each
problem is permutation, combination, or neither:
 1. Number of ways to form a committee of president,
VP, and treasurer from 10 students.
 Permutation
 2. Number of ways to select 5 distinct roles for a play
out of 10 potential actors

14.  Quiz 2.3 #2: Identify whether each
problem is permutation, combination, or neither:
 1. Number of ways to form a committee of president,
VP, and treasurer from 10 students.
 Permutation
 2. Number of ways to select 5 distinct roles for a play
out of 10 potential actors
 Permutation

15.  Quiz 2.3 #2: Identify whether each
problem is permutation, combination, or neither:
 1. Number of ways to form a committee of president,
VP, and treasurer from 10 students.
 Permutation
 2. Number of ways to select 5 distinct roles for a play
out of 10 potential actors
 Permutation
 3. Number of ways to pick a hand of 5 cards from a
deck of cards

16.  Quiz 2.3 #2: Identify whether each
problem is permutation, combination, or neither:
 1. Number of ways to form a committee of president,
VP, and treasurer from 10 students.
 Permutation
 2. Number of ways to select 5 distinct roles for a play
out of 10 potential actors
 Permutation
 3. Number of ways to pick a hand of 5 cards from a
deck of cards
 Combination

17. Choosing
from Multiple Pools
 There are 4 Democrats and 3 Republicans forming a
committee with 2 Democrats and 2 Republicans.
How many different committees can be formed?
 We have to treat the Democrats and Republicans
separately. This is a combination problem. So we
have C(4,2) for Democrats, and C(3,2) for
Republicans. You multiply the two:
C(4,2) x C(3,2) = 6 x 3 = 18
2 Dem out of 4 2 Rep out of 3

18. Choosing
with Multi-Scenarios
 A group of 6 friends are thinking about a Spring break trip to
Florida. At least 4 of them have to go in order to get the group
discount flight. How many groups can be formed such that
they can get the discount?
 This is also a combination problem. For this problem, at least
4 of them have to go, meaning the qualifying events are when
4 or 5 or 6 of them can go. You have to treat each scenario
separately. So we have C(6,4) if 4 of them go, C(6,5) if five go,
and C(6,6) if six go. This time, you add them:
C(6,4) + C(6,5) + C(6,6) = 15 + 6 + 1 = 22
Group of 4 Group of 5 Group of 6

19. Golden Rule of Finite
 Here it is:
AND you multiply, OR you add.
 Let’s take the previous two examples.
 When we need 2 Democrats AND 2
Republicans, we multiplied C(4,2) x C(3,2).
 When we need a group of 4 OR 5 OR 6, we
added
C(6,4) + C(6,5) + C(6,6).

20. Quiz 2.3 #3
 I have 5 dimes and 2 quarters in my pocket. I
reach in and pick 3 coins at random. How many
ways can the 3 coins be selected?
A. 35
B. 70
C. 115

21. Quiz 2.3 #3
 I have 5 dimes and 2 quarters in my pocket. I
reach in and pick 3 coins at random. How many
ways can the 3 coins be selected?
A. 35
B. 70
C. 115
 Answer: A

22. Multiple Pools AND Scenarios
 Brad and Angie have 6 children, 3 boys and 3 girls.
Three of them have to do dishes tonight. How many
ways can the three be selected so that there is at least
one boy and at least one girl?
 First, we’ll draw out a table representing the spectrum of
events, ranging from all from one pool to all from the
other:
0boy 1boy 2boys 3boys
3girls 2girls 1girl 0girl
 Checks represents qualifying events, crosses represents
non-qualifying events

23. Multiple Pools AND Scenarios
 So, 2b1g and 2g1b are the qualifying events.
 Now, since either 2b1g or 2g1b will satisfy the
condition, we add up the number of elements in
those events:
C(3,2)C(3,1) + C(3,1)C(3,2) = 9 + 9 = 18
2boys 1girl 1boy 2girls

24. Complementary Selection
 Sometimes there are more checks than crosses. In
that case, it will be quicker to take advantage of this
equation:
# elements qualify = Total # elements in SS - # elements not quality
 Lets consider the last example, but instead of at
least one boy one girl, we need only at least one
boy:
0boy 1boy 2boys 3boys
3girls 2girls 1girl 0girl

25. Complementary Selection
 In this case, it’s easier if we take total # of
elements in the sample space and subtract the #
in non-qualifying events from it:
C(6,3) – C(3,3)*C(3,0) = 20 – 1= 19
any 3 3girls AND 0boys

26. Quiz 2.3 #4
 I have 6 regular and 4 diet coke in my cooler. I
pick 5 at random. What’s the number of ways I
pick at least 1 diet?
A. 66
B. 246
C. 846

27. Quiz 2.3 #4
 I have 6 regular and 4 diet coke in my cooler. I
pick 5 at random. What’s the number of ways I
pick at least 1 diet?
A. 66
B. 246
C. 846
 Answer: B

28. Summary
 Definition:
 Combination (order doesn’t matter, no repetition)
 How to find the number of combinations
 Formula
 How to distinguish between combination and
permutation
 Problems for multiple pools and multi-scenarios
 Golden rule of Finite

29.  Features
 27 Recorded Lectures
 Over 116 practice problems with recorded solutions
 Discussion boards/homework help
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