Permutations - Exercise set 1

This exercise set contains some solved exercises on permutations. The theory needed to solve these exercises is introduced in the lecture entitled Permutations.

Exercise 1.1

There are 5 seats around a table and 5 people to be seated at the table. In how many ways can they seat themselves?

Solution

Sitting 5 people at the table is a sequential problem. We need to assign a person to the first chair. There are 5 possible ways to do this. Then we need to assign a person to the second chair. There are 4 possible ways to do this, because one person has already been assigned. An so on, until there remain one free chair and one person to be seated. Therefore, the number of ways to seat the 5 people at the table is equal to the number of permutations of 5 objects (without repetition). If we denote it by $P_{5}$ , then: [eq1]

Exercise 1.2

Bob, John, Luke and Tim play a tennis tournament. The rules of the tournament are such that at the end of the tournament a ranking will be made and there will be no ties. How many different rankings can there be?

Solution

Ranking 4 people is a sequential problem. We need to assign a person to the first place. There are 4 possible ways to do this. Then we need to assign a person to the second place. There are 3 possible ways to do this, because one person has already been assigned. An so on, until there remains one person to be assigned. Therefore, the number of ways to rank the 4 people participating in the tournament is equal to the number of permutations of 4 objects (without repetition). If we denote it by $P_{4}$ , then: [eq2]

Exercise 1.3

A byte is a number consisting of 8 digits that can be equal either to 0 or to 1. How many different bytes are there?

Solution

To answer this question we need to follow a line of reasoning similar to the one we followed when we derived the number of permutations with repetition. There are 2 possible ways to choose the first digit and 2 possible ways to choose the second digit. So, there are 4 possible ways to choose the first two digits. There are 2 possible ways two choose the third digit and 4 possible ways to choose the first two. Thus, there are 8 possible ways to choose the first three digits. An so on, until we have chosen all digits. Therefore, the number of ways to choose the 8 digits is equal to: [eq3]

by Marco Taboga

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