Permutation & Combination
Permutation & Combination is an important topic for various competitive exams like IBPS SO, IBPS PO, SBI PO, SBI Clerk, SEBI Grade A, RBI Grade B. Therefore, it is essential to clear the basics of this chapter if you are appearing for any of the above mentioned exams.
Definition of Permutation & Combinations
In Permutation & Combinations we look at various ways in which characters from a given set can be arranged to form subsets without replacements.
In Permutation, an orderly arrangement of elements of a set is involved. Whereas in Combination, we need to look for a number of ways a given set of characters can be arranged without considering their order.
Difference between Permutations and Combinations
Sometimes, the words permutation and combination are used interchangeably. However, they have different implications.
Let’s look at a simple example to understand-
The password of your computer is 1234, but if you enter 4321, it will not unlock even though the numbers are the same. What your computer will recognise is the order of the numbers. There are many possible combinations for the given set of numbers, but your computer accepts only a specific permutation.
This is the basic difference between the two terms.
| Difference | Permutation | Combination |
|---|---|---|
|
Application |
Arranging numbers, digits, alphabets, colours, people |
Selection of teams, food, clothes |
|
Order |
Order matters |
Order does not matter |
|
Used for |
Lists |
Groups |
|
Denoted by |
nPr |
nCr |
Permutation
Factorial
Before getting into the basics of this chapter, let us understand what a ‘factorial’ is.
The product of the numbers starting from 1 up to a number ‘n’ is known as the factorial number of ‘n’.
Meaning, n!= 1x2x3x4x5x6…….x(n-2)x(n-1)xn
1!= 1
2!= 1x2= 2
3!= 1x2x3= 6
4!= 1x2x3x4=24
Note1- 0! and 1! Are equal to 1.
Note 2- We cannot find a factorial of a negative number.
Application of factorial
Factorial is most commonly used in arrangements.
For example- We need to arrange 5 persons in a single line. So, we will start with the first place, which means that we can choose 1 person out of the 5 for the first place. This can be done in 5 ways.
Now, 4 places are vacant and 4 people are left. So now, we can choose 1 person out of the 4 for second place. This can be done in 4 ways.
We will repeat the same process for the other 3 places.
To get the final answer we will multiply all these ways for getting the different ways of arrangement.
Therefore, total ways= 5x4x3x2x1 which is 5!= 120
Q1) In how many ways can the letters of the word PATNA be rearranged?
Answer: PATNA has a total of 5 words. Therefore, we will arrange 5 letters in 5 places in 5!= 120
However, in this question the letter A has been repeated twice. So, we have to divide by the number of repetitions of the word when any letter appears more than once. In this case there are 2 repetitions.
This means, we have to divide the total 120 ways by 2!= 2
So, the total number of arrangements that can be made= 120/2= 60
There is a way to solve this directly as well= 5!/2!
Combinations
Combinations are relatively easier to solve as the order does not matter here. We can select things at random and check out the different possibilities. Therefore, it is a one step process.
Formula for Combination is nCr= n!/r!* (n-r)!
Let’s look at an example to understand-
In how many ways can a coach choose three players from among five players?
Answer: There are 5 players to be taken 3 at a time.
Using the formula:
C(5,3) = P(5,3)/ 3!
= 5×4×3/3×2×1
=10
Thus, the coach can choose the players in 10 different ways.
Formulas for Permutation & Combination
In order to solve question on permutation & combination easily, you need to remember the formulas-
| Permutation or Combination | Repetition | Formula |
|---|---|---|
|
Permutation |
Yes |
P (n,r) = nr |
|
Permutation |
No |
P (n,r) = n! / (n – r)! |
|
Combination |
Yes |
C (n,r) = n! / r! ( n-r)! |
|
Combination |
No |
C (n+ r -1 ,r) = (n + r -1 )! / r! (n – 1) ! |
Solved Questions
Q1) In a class there are 4 boys and 5 girls. In how many ways can a class monitor can be chosen?
Answer: Here, we have to choose 1 student out of 9 to be the monitor.
So, as per the formula nCr= 9C1= 9/1=9
Q2) How many different words can be made using letters of PATNA starting with P?
Answer: PATNA has 5 words. As per the question, P is fixed in the first place. So, we need to arrange the remaining 4 letters at 4 places= 4!= 24 ways.
However, the letter A is repeated twice, so we need to divide the total 24 ways by 2!= 2.
Therefore, different words starting from P= 24/2= 12
Direct approach= 4!/2!= 12
Q3) In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
Answer: The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA).
Hence we can assume total letters as 5 and all these letters are different.
Number of ways to arrange these letters= 5!= 5x4x3x2x1= 120
All the 3 words (OIA) are different
Number of ways to arrange these vowels among themselves= 3!= 3x2x1= 6
Hence, required number of ways= 120x6= 720
Q4) An urn contains 5 red balls and 3 blue balls. In how many different ways can 2 red and 1 blues balls be drawn?
Answer: Ways of selecting 2 red balls= 5C2= 10
Similarly, ways of selecting 1 blue ball= 3C1= 3
So total ways to select 2 red and 1 blue ball= 10*3= 30
