Permutations and Combinations | A Level Maths Revision Notes

Contents hide

1 Summary

1.1 What is Permutation?

1.2 Example #1

1.3 What is Combination?

1.4 Example #2

1.4.1 Reference

Summary

Permutations are ways in which an element from a set is selected considering that the order is significant
${ P }_{ r }^{ n }\quad =\quad \frac { n! }{ (n\quad -\quad r)! }$ when repetition is not allowed
${ n }^{ r }$ when repetition is allowed
Combinations are defined as ways in which an element from a set is selected, such that the order of selection does not matter
${ C }_{ r }^{ n }\quad =\quad \frac { n! }{ r!(n\quad -\quad r)! }$ when repetition is not allowed
$\frac { (n\quad +\quad r\quad -1)! }{ r!(n\quad -\quad 1)! }$ when repetition is allowed

All the above formulas are defined for Number of Permutations or Combinations of r objects chosen from n objects .

Both permutations and combinations are groups or arrangements of objects.. When dealing with combinations, the order of the objects is insignificant, whereas in permutations the order of the objects makes a difference. Permutations are studied in almost every branch of mathematics. They are also used in computer science t for analyzing and sorting algorithms, in quantum physics for describing states of particles and in biology for describing RNA sequences. This articles explains both permutation and combination in detail.

What is Permutation?

Permutation is important for sets where order matters. It is defined as ways in which an element from a set is selected considering that the order is significant. For example, if we are given a word ”HOUSE”, and we need to find different ways in which the letters of this word can be arranged, we first see that the order of letters in this word are important as when we we write the letters in a different order like ”UESOH” it is a completely different arrangement and makes a difference to the word, so we can see that the order matters here. Therefore we will use permutation formulas to solve this.

To work out permutations, given formulas are:

Number of Permutations of r objects from n objects when no repetition is allowed is defined as:

${ P }_{ r }^{ n }\quad =\quad \frac { n! }{ (n\quad -\quad r)! }$

where $n!\quad =\quad n\quad \times \quad (n\quad -\quad 1)\quad \times \quad (n\quad -\quad 2)\quad \times \quad .....\quad 1$

Number of Permutations of r objects from n objects when repetition is allowed is defined as:

${ n }^{ r }$

Example #1

Q. If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this be done?

Solution:

We have to find number of permutations of 4 objects out of 6 objects.

Hence we know, r = 4 and n = 6

As repetitions are not allowed here, we will use the formula:

${ P }_{ r }^{ n }\quad =\quad \frac { n! }{ (n\quad -\quad r)! }$

${ P }_{ 4 }^{ 6 }\quad =\quad \frac { 6! }{ (6\quad -\quad 4)! }$

${ P }_{ 4 }^{ 6 }\quad =\quad \frac { 6\quad \times \quad 5\quad \times \quad 4\quad \times \quad 3\quad \times \quad 2\quad \times 1 }{ 2\quad \times \quad 1 }$

${ P }_{ 4 }^{ 6 }\quad =\quad 360$

Therefore, we can say that cards can be sent in 360 different ways.

What is Combination?

The combination is defined as ways in which an element from a set is selected, such that (unlike permutations) the order of selection does not matter.

For a combination, the set ABC is similar to BCA. They are used for groups where you have to select items from a collection and order can be decided by yourself. An example of this would be Think of combining ingredients flour, salt and water in a bowl is the same as salt, water and flour. As it does not matter in which order these ingredients are added here.

To work out combinations, given formulas are:

Number of Combinations of r objects chosen from n objects when no repetition is allowed is defined as:

${ C }_{ r }^{ n }\quad =\quad \frac { n! }{ r!(n\quad -\quad r)! }$

Number of Combinations of r objects chosen from n objects when repetition is allowed is defined as:

$\frac { (n\quad +\quad r\quad -1)! }{ r!(n\quad -\quad 1)! }$

Example #2

Q. Five bingo numbers are being picked from a ball containing 100 bingo numbers. How many possible ways are there for picking different numbers?

Solution:

Since the order does not matter here, we know its combinations and from the question we extracted relevant data r = 5 and n = 100

Put these values in the above given formula to find out the combinations:

${ C }_{ r }^{ n }\quad =\quad \frac { n! }{ r!(n\quad -\quad r)! }$

${ C }_{ 5 }^{ 100 }\quad =\quad \frac { 100! }{ 5!(100\quad -\quad 5)! }$

${ C }_{ 5 }^{ 100 }\quad =\quad 75,287,520$

Therefore, we worked out there are 75,287,520, different ways for picking different numbers.