Download 13-1 Permutations and Combinations

Advanced Mathematical Concepts
Chapter 13
Lesson 13-1
Example 1
Chelsea is deciding what meal to prepare for dinner. She needs to choose a meat, a starch, a vegetable,
and a dessert. For a meat she can choose from beef, chicken, or fish, for a starch she may select a
potato or rice, and for a vegetable she can choose broccoli, string beans, and asparagus. Her dessert
choices are apple pie, peach pie, or blueberry pie.
a. Are the choices for each category of food independent or dependent events?
Since the choice of one type of food does not affect the choice of another type of food, the events are
independent.
b. How many different ways can Chelsea choose the foods for her meal?
Chelsea has three choices for a meat, two choices for a starch, three choices for a vegetable, and three
choices for a dessert.
Meat
3

Starch
2

Vegetable
3

Dessert
3
This can be represented as 3  2  3  3 or 54 different meal choices.
There are 54 possible ways for Chelsea to prepare her meal.
Example 2
Four ice-skaters, Allison, Francesca, Madison, and Alexa are skating in a competition.
a. Are the selections of skaters to perform first, second, third and fourth independent or dependent
events?
b. Assuming that no two skaters may perform at the same time, how many ways can the skaters be
arranged to perform?
a. The choice of skater to perform first does affect the choice of skater to perform second, third, and fourth.
Therefore, the events are dependent.
Advanced Mathematical Concepts
Chapter 13
b. Since order is important, this situation is a permutation.
Method 1: Tree diagram
There are four possibilities for the first performer, three for the second, two for the third, and one for the
fourth as shown in the tree diagram below. If Allison performs first, then Francesca, Madison, or Victoria
will perform second. If Francesca performs second, then either Madison or Victoria will perform third. If
Madison performs third, then Victoria must perform fourth.
There are 24 possible ways that the skaters can be arranged to perform.
Advanced Mathematical Concepts
Chapter 13
Method 2: Permutation Formula
This situation depicts four objects taken four at a time and can be represented as P(4, 4).
P(4, 4)
= 4!
= 4  3  2  1 or 24
Thus, there are 24 ways the skaters can be arranged to perform.
Example 3
In a baseball game, nine players on a team play the field during any given inning. Assume the same
nine players play during the entire game.
a. How many different ways can the coach arrange the batting order?
b. In how many ways can the coach select who will play the outfield positions, that is, left field, center
field, and right field?
a. Since order is important, the situation is a permutation. Also, the 9 players are being taken all at once so
the situation can be represented as P(9, 9).
P(9, 9) = 9!
= 9  8  7  6  5  4  3  2  1 or 362,880
There are 362,880 ways that the coach can arrange the 9 players to bat.
b. This is a permutation of 9 people being chosen 3 at a time.
9!
P(9, 3) = (9 - 3)!
9  8  7  6!
=
6!
= 504
There are 504 ways in which the positions can be filled.
Advanced Mathematical Concepts
Chapter 13
Example 4
CINEMATOGRAPHY In 1998, 195 theatrical films were released in the United States. Of these, only
five were nominated for the academy award for best picture. How many groups of five movies could
have been chosen to be nominated for best picture?
Since order is not important, the selection is a combination of 195 movies selected 5 at a time. It can be
represented as C(195, 5).
195!
(195 - 5)! 5!
195!
= 190! 5!
195  194  193  192  191  190!
=
190! 5!
195  194  193  192  191
=
54321
= 2,231,243,664
C(195, 5) =
There were 2,231,243,664 possible groupings of movies.
Example 5
At Kennedy High School there are 12 names on the ballot for sophomore class officers. Six will be
selected to form a class committee.
a. How many different committees of 6 can be formed?
b. In how many ways can a committee of 6 be formed if each student has a different responsibility?
c. If there are 7 girls and 5 boys on the ballot, how many committees of 3 girls and 3 boys can be
formed?
a. Order is not important in this situation, so the selection is a combination of 12 people chosen 6 at a time.
12!
C(12, 6) = (12 - 6)! 6!
12!
= 6! 6!
12  11  10  9  8  7  6!
=
or 924
6! 6!
There are 924 different ways to form the committees of 6.
Advanced Mathematical Concepts
Chapter 13
b. Order has to be considered in this situation because each committee member has a different
responsibility.
P(12, 6)
12!
= (12 - 6)!
12!
= 6! or 665,280
There are 665,280 possible committees.
c. Order is not important. There are three questions to consider.
How many ways can 3 boys be chosen from 5?
How many ways can 3 girls be chosen from 7?
Then, how many ways can 3 boys and 3 girls be chosen together?
Since the events are independent, the answer is the product of the combinations C(5, 3) and C(7, 3).
7!
(7 - 3)! 3!
5!
7!
= 2! 3!  4! 3!
= 10  35 or 350
C(5, 3)  C(7, 3) =
There are 350 possible committees.