Permutations and Combinations

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.

Topics in Permutations and Combinations