# Permutations and Combinations

Sub Topics
In daily life, we use permutations and combinations a lot. For example, we want to go from Newyork to New Jersey and return by a different mode of transport, out of 16 cricket probables, there are a number of ways of choosing final eleven. There are a number of ways of choosing preliminary and main subjects for civil services exam, there are many ways guests can be seated on a dining table and so on. We will try to learn the basics of Permutations and Combinations. We have covered the same in the pre algebra as well.

### Permutations and Combinations Formula

Permutation Formula is npr= n!/n-r!

Where n = Total number of objects
r = The number of ways to be selected.

Combination Formula is npr= n!/n-r! r!

Where n = Total number of objects
r = The number of ways to be selected.

## Basics of Permutations and Combinations

1) Permutation mean arrangement.

2) Combination mean selection.

3) Fundamental principle of Counting:
a) Multiplication Rule: If a work is done only when all of a number of work are done then number of ways of doing that work is equal to the product of number of way of doing separate work.

b) Addition rule: If a work is done only when any one of a number of work is done , then number of way of doing that work is equal to sum of number of work of way of doing separate work.

4) n! = 1.2.3………………..n.

0! =1

5) Number of permutation of n different things taken r at a time is denoted by n!/n-r!
a) Number of permutation of n different things = n!

b) Number of permutation of n things, out of which p are alike and are of one type, q are alike and are of second type and rest are all different n!/p!q!

## Define Permutations and Combinations

Permutation:

Each of the different arrangements which can be made by taking some or all of a number of given things or object at a time is called a permutation. In permutation order of appearance of things is taken into account. The following examples are taken from algebra problems.

Permutations Examples: The following six arrangement can be made with three distinct object taking two at a time ab, bc, cb, ba, ac, ca. Each of these arrangement is called a permutation.

Combination:

Each of the different groups or selections which can be made by taking some or all of a number of given things or object at a time is called a combination. In combination order of appearance of things is not taken into account.the following algebra answer examples explain this better.

Combination Examples: Following three groups can be made with three different objects taken two at a time. ab, bc, ca here ab and ba are the same group.

## Permutations and Combinations Problems

Below you could see Permutations and Combinations Examples

### Solved Examples

Question 1: In how many ways can 10 students be arranged in 3 seats?
Solution:
We know that the formula for arranging n items taking r at a time is given by

P(n, r) = $\frac{n!}{(n-r)!}$

= n x (n - 1)...(n - r + 1)

Here n = 10 and r = 3

The required permutation = P(10, 3) = $\frac{10!}{(10-3)!}$ = $\frac{10!}{7!}$ = $\frac{3628800}{5040}$ = 720 ways

The number of ways 10students can be arranged in 3 seats = 210

Question 2: Find the number of permutations that can made with the letters of the word “ JANET” taking 4 at a time.
Solution:
We know that the formula for arranging n items taking r at a time is given by

P(n, r) = $\frac{n!}{(n-r)!}$

= n x (n-1)...(n - r + 1)

Here there are 4 letters in the given word.  So n = 5.

Here we have to find the permutations taking 4 at a time.  So r = 4

So required permutation is P(5, 4) = 5 x 4 x 3 x 2 =  24

Therefore the permutations that can made with the letters of the word “ JANET” taking 4 at a time = 24

Question 3: There are 15 candidates for 4 vacancies. In how many ways can they be selected?
Solution:
We know that the formula for selecting n items taking r at a time is given by
C(n,r) = $\frac{n!}{[(n-r)!r!]}$

= [n x (n-1)...(n-r+1)]/[1 x 2 x ...r]

Here n =15 and r = 4

The required combination = C(15, 4) = $\frac{15!}{[(15 - 4)!4!]}$ = $\frac{15!}{11!}$ x 4! = $\frac{1307674368000}{(39916800 \times 24)}$ = 1365

The number of ways selecting 15 candidates for 4 vacancies = 1365

## Permutations and Combinations Practice

Below you could see permutations and combinations practice problems

### Practice Problems

Question 1: Find the number of ways of arranging 10 different objects taken 2 at a time.
Question 2: In how many ways can you select a bead colored green of 5 from a total of 12 different colored balls.