Section 4.8: Combinations
In the last section we solved problems in which
the order that the objects are arranged was
important. We used the Counting Principle and
a Permutation formula to answer the questions
that were asked. The permutation formula and
the Counting Principle over count when the
order that the objects we select are placed is
not important.
Order is not always important. For instance if I
purchase an Arizona Pick Lottery ticket I only
need to match the numbers drawn to win and
the order the numbers were drawn does not
affect whether I win or not. If I match all 6
numbers, regardless of order I win.
Example: Holly, Sarah, Jen and Marsha are at a
Zac Brown concert. A stage hand comes up to
the group and tells them that 3 of them can go
backstage to meet the band. How many
different groupings are possible?
If you used the Counting principle you would
make three slots, fill the slots with numbers
and multiply to get 24 different groups.
____ * ____ * ____
1st
2nd
3rd
4*3*2 = 24
If you tried to list all of the groups possible, you
would run out of groups long before you get to
24. In fact there are only 4 groups possible.
Basically each group picks 3 girls and leaves 1
girl out. It’s easy to make the list if you just
leave a different girl out each time you make an
entry to the list.
Here are the 4 possible groups:
Holly, Sarah, Jen
Holly, Sarah, Marsha
Holly, Jen, Marsha
Sarah, Jen, Marsha
Any other group you might list would just be
one of these groups with the names
rearranged.
The counting principle over counts when only
the objects selected is important and the order
that they are selected in is not important.
When the order of selection is not important
the problem is a COMBINATION problem. Here
is a definition of the topic we cover in this
section – Combination.
Combination is a way of selecting members
from a grouping, such that the order of
selection does not matter.
When order is not important we can’t just
make lines for each choice and put numbers
over the lines and multiply. This will over count
when order is not important.
We will need a formula to find out how many
combinations (arrangements in which order is
not important) from a group are possible.
The formula we need:
n
Cr 
n!
 n  r !r !
n = total number of objects
r = number being selected
The Zac Brown problem could be solved using
this formula. In this case the n = 4, as the group
has 4 people and the r = 3 as only 3 of them can
go backstage.
4C3 =
(
4!
4−3)!3!
=(
4∗3∗2∗1
1)!∗3∗2∗1
=
4∗3∗2∗1
1∗3∗2∗1
there are 4 possible groups)
= 4 (thus
Example: How many different ways can three
people from Dave, Dan, Julie, Amanda and
Steve be selected to attend a meeting?
Before putting pencil to paper, I need to ask
myself if the order the people are selected is
important or not. If I think the order is
important then it is a permutation problem and
I can use the permutation formula, or make
lines and multiply. If I think order is not
important then I need to use the combination
formula.
That is if you think picking Dan, Dave and Julie
to attend meeting is different the picking Dave,
Julie and Dan to attend the meeting you would
solve this problem using the techniques from
section 4.7. I don’t think this would be correct.
The order you list the people selected to attend
a meeting is likely not important. This is
supposed to be a combination problem. I will
solve it using the combination formula.
5 C3 
5!

5

3
!3!


Answer: 10
Example: 10 GCC students have applied for a
scholarship. 6 students will be chosen to
receive this scholarship, how many different
ways can these 6 be chosen?
Because the order that you were selected
doesn’t matter, and only the fact that you are
selected matters this is a combination problem.
10 C6 
10!
(10−6)!6!
Answer: 210
Homework #1-17
1) An ice cream parlor has 20 different flavors.
Cynthia orders a banana split and has to select
3 different flavors. How many different
selections are possible?
2) A textbook search committee is considering
8 books for possible adoption. The committee
has decided to select three of the eight for
further consideration. In how many ways can
they do so?
3) The drama club is planning a trip to New
York. While in New York they want to attend 3
plays out of 10 plays they would like to see.
How many different groups of 3 plays can they
select to attend?
4) Neo wants to download 6 different songs.
He only has enough money to purchase 4. In
how many ways can he select 4 or 6 songs to
purchase?
5) In how many ways can you select a
committee of 4 students out of 12 students?
6) In how many ways can you select 3 different
vegetables from 8 vegetables?
7) 3 different varieties of flowers are going to
be placed in a vase. There are 5 different
flowers to choose from. How many
combinations can be made from the 5 varieties
of flowers?
8) How many 3 topping pizzas could be made if
you have 10 toppings to choose from (assume
each topping can only be used once)?
9) The US senate consists of 100 senators, two
from each state. A committee consisting of five
senators is to be formed. How many are
possible?
10) There are 6 restaurants you would like to
try on your next vacation. You only have time
to try 4 of them when you go on vacation. How
many groups of restaurants can you choose?
11) You need to upload a few videos to watch
on a plane. You only have storage capacity to
add 5 new videos, but have 12 videos to choose
from. How many combinations videos can you
choose?
12) How many groups of 2 sevens can be made
from the 4 sevens in a deck of cards?
13) How many groups of 3 kings can be made
from the 4 kings in a deck of cards?
14) How many groups of 4 hearts can be made
from the 13 hearts in a deck of cards?
15) How many groups of 3 spades can be made
from the 13 spades in a deck of cards?
16) How many groups of 2 black cards can be
made from the 26 black cards in a deck of
cards?
17) How many groups of 5 red cards can be
made from the 26 red cards in a deck of cards?
There are times when it is okay and necessary
to use both the Counting Principle and the
Combination formula to solve a problem. The
following example is a blended problem that
uses the combination formula along with the
counting principle.
Example: At a Chinese restaurant, dinner for 8
people consists of 3 items from column A,
4 items from column B and 3 items from
column C. If columns A, B and C have 5, 7 and 6
items respectively how many different dinner
combinations are possible?
This is where things get very confusing; I am
using the counting principle for this problem,
along with the combination formula. Knowing
how to combine rules can be tricky. I will use
combinations for the number of choices of each
item, and the counting principle for the total
number of combinations.
Choose 3
of 5 items
from
column A
5C3 = 10
Choose 4
of 7 items
from
column B
7C4 = 35
Choose 3
of 6 items
from
column C
6C3 = 20
Now multiply the results: 10*35*20 = 7000
Answer there are 7000 different combinations
possible.
Example: How many groups of 3 hearts and 5
spades can be made from the 13 hearts and 13
spades in a deck of cards?
Choose 3 of 13 hearts Choose 5 of 13 spades
13C3 = 286
13C5 = 1287
Now multiply the results: 286*1287 = 368,082
Answer: There are 368,082 ways
Homework #18 – 32
19) A couple is planning a party. They need to
pick 2 of 6 dinners to offer, and 3 of 5 desserts.
How many different combinations can they
pick?
18) On an English test, Tito must write an essay
for 3 of 5 questions in part one and 4 of 6
questions in part two. How many different
combinations of questions can he answer?
21) 6 members of a club will be selected to
attend a conference. The club has 8 female and
7 male members. How many different groups
can be selected if 3 male and 3 females must be
selected?
20) An 8 person committee is being formed
from a club that has 10 seniors and 8 juniors.
How many committees can be formed if the
committee must have 6 seniors and 2 juniors?
23) At a medical research center an
experimental drug is to be given to 12 people, 6
men and 6 women. If 10 men and 9 women
have volunteered to be given the drug, in how
many ways can the researcher choose the 12
people to be given the drug?
22) A club plans to visit 3 of 5 landmarks in
California and 2 of 4 landmarks in Oregon. How
many different groups of sites can they select?
25) Michael is sent to the store to get 5
different bottles of regular soda and 3 different
bottles of diet soda. If there are 10 different
types of regular soda, and 7 different types of
diet soda to choose from, how many different
choices does Michael have?
24) At a medical research center an
experimental drug is to be given to 20 people,
12 children and 8 adults. If 15 children and 9
adults have volunteered to be given the drug,
in how many ways can the researcher choose
the 20 people to be given the drug?
27) How many groups of 2 sevens and 3 threes
can be made from the 4 sevens and 4 threes in
a deck of cards?
26) A couple is throwing a party. They need to
pick 5 different kinds of beer from the 12 beers
being offered, and they need to pick 4 snacks
from the 9 snacks being offered. How many
different choices do they have?
29) How many groups of 4 hearts and 2 spades
can be made from the 13 hearts and 13 spades
in a deck of cards?
28) How many groups of 3 kings and 1 jack can
be made from the 4 kings and 4 jacks in a deck
of cards?
31) How many groups of 2 black cards and 4
red cards can be made from the 26 black cards
and 26 red cards in a deck of cards?
30) How many groups of 3 spades and 2 clubs
can be made from the 13 spades and 13 clubs
in a deck of cards?
32) How many groups of 5 red cards and 1
black card can be made from the 26 red cards
and 26 black cards in a deck of cards?
Answers: 1) 20C3 = 1140 3) 10C3 = 120 5)
12C4 = 495 7) 5C3= 10 9) 100C5 = 75,287,520
11) 12C5 =792 13) 4C3 = 4 15) 13C3 =286
17) 26C5 =65,780
19) 6C2 * 5C3 = 15*10 = 150 21) 7C3 * 8C3 =
35*56 = 1960 23) 10C6 * 9C6 = 210*84 =
17,640
25) 10C5 * 7C3 = 252*35 = 8,820 27) 4C3 * 4C2 =
4*6 = 24 29) 13C4 * 13C2 = 715 * 78 = 55,770
31) 26C2 * 26C4 = 325*14950 = 4,858,750