Finding Permutations and Combinations
Algebra II Lesson 42
A factorial of a positive integer is the product
of all positive integers up to and including that
integer. Note that 0 factorial (0!) is defined to
be 1.
n Factorial
The factorial of n is denoted as 𝑛!
𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) . . . 1
Zero factorial is defined to be 1. 0! = 1.
Algebra II Lesson 42
Evaluate the following expressions that contain
factorials:
a. 5!
SOLUTION
5!=5×4×3×2×1 =120
A factorial of a positive integer is the
multiplication of all positive integers up to and
including the integer.
Algebra II Lesson 42
Evaluate the following expressions that contain factorials:
b.
10!
2! (10 − 2)!
SOLUTION
10!
2! (10 − 2)!
10!
2! 8!
To help simplify expressions with factorials, expand the numerator
just enough to cancel duplicate factors in the denominator.
10 × 9 × 8!
2 × 8!
90
= 45
2
Algebra II Lesson 42
A permutation is a selection of items where
order is important.
Permutations
The permutation of n objects taken r at a time
is:
𝑛!
𝑃(𝑛, 𝑟) =
(𝑛 − 𝑟)!
The permutation of 6 objects taken 3 at a time
is:
6!
𝑃(6,3) =
(6 − 3)!
Algebra II Lesson 42
A permutation of n objects arranged into one
group of size n, without repetition of objects, can
be found by using the formula: 𝑃(𝑛, 𝑛) = 𝑛!
For example, the permutations of three letters (A,
B, C) arranged into groups of three can be
calculated as 𝑃(3,3) = 3! = 3 × 2 × 1 = 6. We can
check our calculation by showing all possible
arrangements of the three letters, without
repetition. The 6 arrangements are ABC, ACB, BAC,
BCA, CAB, and CBA.
Algebra II Lesson 42
When letters are repeated, permutations are not
distinguishable from each other. For example, for
the letters (B, B, O) there are only three
distinguishable permutations of the letters: BOB,
BBO, and OBB.
The number of distinguishable permutations of n
objects where one object is repeated 𝑞1 times,
another object is repeated 𝑞2 times, and so on is:
𝑛!
𝑞1 ! ∙ 𝑞2 ! ∙ ⋯ ∙ 𝑞𝑘 !
The number of permutations of the letters BOB is
3!
6
=2=3
2!1!
Algebra II Lesson 42
a. Find the permutation of 8 objects.
SOLUTION
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
The permutation of n objects is n!
There are 40,320 possible permutations of 8
objects.
Algebra II Lesson 42
b. Find the permutation of 6 objects taken 2 at a time.
SOLUTION
Use the formula for the permutation of n objects taken r at a time.
𝑛!
𝑃(𝑛, 𝑟) =
(𝑛 − 𝑟)!
6!
𝑃(6,2) =
(6 − 2)!
6!
𝑃(6,2) =
4!
6 × 5 × 4!
𝑃(6,2) =
4!
𝑃(6,2) = 30
There are 30 possible permutations of 6 objects taken 2 at a time.
Algebra II Lesson 42
c. Find the number of distinguishable permutations of the letters in the
word CONNECTION.
SOLUTION
CONNECTION has 10 letters with C repeated 2 times, O repeated 2 times,
and N repeated 3 times. Let n = number of letters in the word
CONNECTION, 𝑞1 = number of Cs, 𝑞2 = number of Os, and 𝑞3 = number of
Ns.
The number of distinguishable permutations can be found using:
𝑛!
𝑞1 ! 𝑞2 ! 𝑞3 !
10!
2! 2! 3!
3,628,800
2×2×6
151,200
There are 151,200 distinguishable permutations of the letters in the word
CONNECTION.
Algebra II Lesson 42
A combination is a selection of items where order does not matter.
Combinations
The combination of n objects taken r at a time is:
𝑛!
𝐶(𝑛, 𝑟) =
𝑟! (𝑛 − 𝑟)!
The combination of 5 objects taken 2 at a time is:
5!
𝐶(5,2) =
2! (5 − 2)!
5!
𝐶(5,2) =
2! 3!
5 × 4 × 3!
𝐶(5,2) =
2 × 3!
𝐶 5,2 = 5 × 2 = 10
Algebra II Lesson 42
To decide if a permutation or combination is
needed, ask yourself if order is important.
The arrangement of digits for an alarm system
code requires a specific order; the code 352 is
different from the code 235, so a permutation is
used to determine the number of possible codes.
The arrangement of names of people in a study
group does not require a specific order, so a
combination is used to calculate the number of
possible arrangements.
Algebra II Lesson 42
a. Find the combination of 7 objects taken 4 at a time.
SOLUTION
Use the formula for the combination of n objects taken r at a
time. So,
𝑛!
𝐶(𝑛, 𝑟) =
𝑟! (𝑛 − 𝑟)!
7!
𝐶(7,4) =
4! (7 − 4)!
7!
𝐶(7,4) =
4! 3!
7 × 6 × 5 × 4!
𝐶(7,4) =
4! 3 × 2 × 1
35
There are 35 combinations of 7 objects taken 4 at a time.
Algebra II Lesson 42
b. How many different pizza varieties can be created if
there are two choices of crust, thick or thin, and two
out of three possible toppings are selected?
SOLUTION
The combination of the 2 types of crusts is 𝐶(2,1). The
combination of 2 out of 3 toppings is 𝐶(3,2). So, the
number of pizza varieties that can be created is:
𝐶(2,1) × 𝐶(3,2)
2!
3!
×
1! 2 − 1 ! 2! (3 − 2)!
2 3 × 2!
×
1
2!
6
Algebra II Lesson 42
c. To travel from New York to Boston you could take one of three
trains or one of four buses. How many different options for
traveling between the two cities do you have?
SOLUTION
The combination of trains is 𝐶(3,1). The combination of buses is
𝐶(4,1).
So the number of travel options is 𝐶(3,1) + 𝐶(4,1)
𝐶 3,1 + 𝐶 4,1
3!
4!
+
1! 3 − 1 ! 1! 4 − 1 !
3! 4!
+
2! 3!
3 × 2! 4 × 3!
+
2!
3!
3+4 = 7
Algebra II Lesson 42
a. Explain whether the following is a combination or a
permutation and find the solution. How many twoletter arrangements can be formed from the letters
CAT?
SOLUTION
Order makes a difference in this problem because CA
and AC are not the same arrangement.
The solution is found using permutations:
3!
𝑃 3,2 =
3−2 !
6
There are 6 possible arrangements
{CA, AC, CT, TC, AT, TA}.
Algebra II Lesson 42
b. Explain whether the following is a combination or a
permutation and find the solution. How many two-man
crews can be selected from the set of three males:
{Charlie, Anton, Tom}?
SOLUTION
For this problem, order does not make a difference.
The two-man crew of Charlie/Anton is the same as the
two-man crew Anton/Charlie.
The solution is found using combinations:
3!
𝐶 3,2 =
2! 3 − 2 !
3
There are only 3 different two-man crews
{Charlie/Anton, Charlie/Tom, Anton/Tom}.
Algebra II Lesson 42
By labeling the top row as Row 0 and the first entry in each row as Entry 0,
Pascal’s Triangle shows a direct relationship to combinations. For example,
to count the number of ways to choose 1 object from 3 objects, look at
Entry 1 in Row 3. This entry is 3. 𝐶(3,1) = 3
Explore the different combinations represented in Pascal’s Triangle by
identifying all of the combinations in the first 5 rows. Justify your results by
solving for each combination identified.
SOLUTION
Combinations:
𝐶(0,0) = 1;
𝐶(1,0) = 1; 𝐶(1,1) = 1;
𝐶(2,0) = 1; 𝐶(2,1) = 2; 𝐶(2,2) = 1;
𝐶(3,0) = 1; 𝐶(3,1) = 3; 𝐶(3,2) = 3; 𝐶(3,3) = 1;
𝐶(4,0) = 1; 𝐶(4,1) = 4; 𝐶(4,2) = 6; 𝐶(4,3) = 4; 𝐶(4,4) = 1;
𝐶(5,0) = 1; 𝐶(5,1) = 5; 𝐶(5,2) = 10; 𝐶(5,3) = 10; 𝐶(5,4) = 5; 𝐶(5,5) = 1.
Algebra II Lesson 42
A school has scheduled 5 football games, 6 basketball games, 3 volleyball games, and 4 soccer games. You have a
ticket that allows you to attend 5 games. In how many ways can you attend 2 football games, 2 basketball games,
and either a volleyball or soccer game?
SOLUTION
1. Understand: Since the order of arrangements does not matter, this is a combination problem.
2. Plan: To find the number of ways to use your tickets, make a table to help solve the problem.
Five-Game Ticket
Attend 2 games
Attend 2 games
Attend 1 game
Football: 5
Basketball: 6
Volleyball or soccer: 7
Use combinations to solve. Multiply the combination for each event to find the possible ways to use your tickets.
3. Solve: 𝐶(5 𝑓𝑜𝑜𝑡𝑏𝑎𝑙𝑙, 𝑝𝑖𝑐𝑘 2) × 𝐶(6 𝑏𝑎𝑠𝑘𝑒𝑡𝑏𝑎𝑙𝑙, 𝑝𝑖𝑐𝑘 2) × (7 𝑣𝑜𝑙𝑙𝑒𝑦𝑏𝑎𝑙𝑙 𝑜𝑟 𝑠𝑜𝑐𝑐𝑒𝑟, 𝑝𝑖𝑐𝑘 1)
𝐶 5,2 × 𝐶 6,2 × 7,1
5!
6!
7!
×
×
2! 5 − 2 ! 2! 6 − 2 ! 1! 7 − 1 !
5 × 4 × 3! 6 × 5 × 4! 7 × 6!
×
×
2 × 3!
2 × 4!
6!
10 × 15 × 7
1,050
Check: Use your calculator to check your calculations for each combination, and then multiply the results.
5𝐶2 = 10; 6𝐶2 = 15; 7𝐶1 = 7;
10 × 15 × 7 = 1,050
Algebra II Lesson 42
a. Evaluate 6!
b. Evaluate
6!
4!(6−4)!
Algebra II Lesson 42
c. Find the number of permutation of 7 objects.
d. Find the permutation of 12 objects taken 3
at a time.
Algebra II Lesson 42
e. OHIO has 4 letters with O repeated twice.
Find the number of distinguishable
permutations of letters in OHIO.
f. Find the combination of 10 objects taken 7
at a time.
Algebra II Lesson 42
g. Find the number of salad varieties using one
of four salad dressings and two of four
additional vegetables.
h. How many different 4-card hands are
possible if you draw 4 cards from a standard
deck of 52 cards? Is this a permutation or
combination problem?
Algebra II Lesson 42
i. How many different ways can 4 cards be
picked from a standard deck of 52 cards and
be played in a game? Is this a permutation or a
combination problem?
j. In how many ways can 3 blue, 2 red, and 5
green markers be distributed to 10 students?
Algebra II Lesson 42
Page 309
Lesson Practice (Ask Mr. Heintz)
Page 309
Practice 1-30 (Do the starred ones first)
Algebra II Lesson 42