M5 2.3 Permutations

Multiple Choice

A teacher, a nurse, and an engineer walk into a restaurant. If there are 3 seats available, how many ways are there for for the teacher, nurse, and engineer to seat themselves?

• 1

• 3

• 6

• 27

Multiple Choice

A teacher, a nurse, an engineer, and a doctor walk into a restaurant. If there are 4 seats available at the restaurant, how many ways are for the teacher, nurse, engineer, and doctor to seat themselves?

• 1

• 4

• 10

• 24

• 256

Multiple Choice

The mathematician, physicist, engineer, and computer scientist walk into a restaurant and decide to seat at a circular table. They file to the table, staying in order as they sit down and wind up seated as shown. How many different seating arrangements around the table in the image above?

Note: "Staying in order" means that each person has the same left and right neighbors that they had when they walked in.

• 0

• 1

• 4

• 6

Multiple Choice

How many different ways can you stack a scoop of chocolate ice cream, a scoop of pistachio ice cream, a scoop of vanilla ice cream, and a scoop of strawberry ice cream onto an ice cream cone?

• 24

• 12

• 16

• 4

Multiple Choice

Six friends want to form a club. They decide that there will be 1 president, 1 secretary and 4 ordinary members.

How many different ways can they organize this club?

• 180

• 720

• 30

• 15

​Six friends want to form a club. They decide that there will be 1 president, 1 secretary and 4 ordinary members. How many different ways can they organize this club?

Multiple Choice

Five children (Ice, Feem, Fame, Non, and Jane) are playing a game of hide-and-go-seek. Ice counts to 100 and the other four children each go to hide in one of the rooms of the house. If there are 7 rooms that the children could hide in, and each hides in a different room, how many different ways can the children hide?

• 5040

• 840

• 210

• 35

​Five children (Ice, Feem, Fame, Non, and Jane) are playing a game of hide-and-go-seek. Ice counts to 100 and the other four children each go to hide in one of the rooms of the house. If there are 7 rooms that the children could hide in, and each hides in a different room, how many different ways can the children hide?

​We can consider this as assigning 4 rooms from a set of 7 to the 4 children.

​There are 840 ways the children could hide.

Multiple Choice

On a trip to the local zoo, there are 6 exhibits that you want to visit. In how many different orders can you visit the exhibits?

• 720

• 360

• 120

• 30

Multiple Choice

There are 3 flights from California to France, and 2 flights from France to India. Tiger wants to fly from California to France and then to India.

How many choices does he have for his flight plan?

• 3

• 8

• 2

• 6

Multiple Choice

You have 3 different-colored bottles, each with a distinct cap.

In how many ways can these caps be put on the bottles such that none of the caps are on the correct bottles?

Assume that all the caps must be on the bottles.

• 1

• 2

• 3

• 6

Multiple Choice

There are 10 students in a class. Two of these students are Jack and Daniel, who don’t get along very well. In how many ways can the teacher arrange the students in a row, so that Jack and Daniel are not together?

• 9!

• 10! - (2 × 9!)

• 10! - 9!

• 9! * 2

​There are 10 students in a class. Two of these students are Jack and Daniel, who don’t get along very well. In how many ways can the teacher arrange the students in a row, so that Jack and Daniel are not together?

​The total number of arrangements of the 10 students is 10P10 = 10!.

Treating Jack + Daniel as a single unit (call it JD), we have a total of 9 entities which we can permute: the 8 students, and JD. These 9 entities can be permuted in 9P9= 9! ways.

​J and D can be permuted among themselves in 2! or 2 ways: JD or DJ. Thus, the total number of permutations in which Jack and Daniel are together, is 2 × 9!

​The the number of permutations in which Jack and Daniel are not together is 10! - (2 × 9!).