Important Info
Today I’m going to discuss a very important topic for ibps or other bank exams i.e. permutation and combination. This is primarily because various questions from this section tests candidate’s analytical skill. This topic just involves basic calculations.
Permutation implies arrangement where order of things is important and includes word formation, number formation, circular permutation etc. Combination means selection where order is not important and it involves selection of team, forming geometrical figures, distribution of things etc.
Factorial = Factorial are defined for natural numbers, not for negative numbers.
n! = n.(n-1).(n-2)………3.2.1
For example: 1) 4! = 4.3.2.1 = 24
2) 6!/ 4! = (6.5.4!)/ 4! = 6.5 = 30
3) 0! = 1
| PERMUTATION |
COMBINATION |
| Implies Arrangement |
Implies Selection |
| Order of things is important |
Order of things is NOT important |
|
Permutation of three things a, b and c taking two at a time are ab, ba, ac, ca,bc and cb (Order is important).
|
Combination of three things a,b and c taking two at a time are ab, ca and cb (Order is not important). |
| nPr= n!/ (n-r)! |
nCr = n!/ (n-r)! r! |
| nPn = n! |
nCn = 1 |
| nP0 = 1 |
nC0 = 1 |
Example of Word Formation
Example: How many new words can be formed with the word “PATNA”?
Solution:
In word “PATNA”, P,T,N occurs once and A occurs twice.
****Always remember in word formation, if word repeats, number of repetition will be on denominator.
So, total number of words that can be formed = 5!/ 2! = 60
Therefore, except PATNA there are 59 new words (60-1).
Example: How many words can be formed from the letters of the word “EXAMINATION”?
Solution:
E, X, M, T, O : Occurs ONCE
A, I, N : Twice
So, total number of words = 11! / 2! 2! 2!
(Total number of letters=11 and 3 letters are occurring twice)
Problems for practice
Problem 1: Choose permutation or combination
1) Selection of captain and bowler for a play.
Permutation
2) Selection of four students for a lecture.
Combination
3) Assigning people to their seats during conference.
Permutation
Problem 2: Evaluate 7P2 . 4P3
Solution:
(7!/ 5!). (4!/ 1!)
⇒(7.6). (4.3.2)
⇒1008
Problem 3: Evaluate 5C2. 3C2
Solution:
(5!/3!2!). (3!/2!1!)
⇒(5.4/2). (3)
⇒30
Problem 4: How many ways are there in selecting 5 members from 6 males and 5 females, consisting 3 males and 2 females?
Solution:
This is a case of combination i.e.selecting 3 males from 6 males and 2 females from 5 females.
⇒Required number of ways = (6C3 *5C2)
⇒(6.5.4/3.2)*(5.4/2)
⇒200.
Problem 5: How many words can be formed by using letters of the word “DAUGHTER” so that the vowels come together?
Solution:
This is a case of permutation. In a word “DAUGHTER”, there are 8 letters including 3 vowels (AUE)
According to the question, vowels should always come together. Therefore, in this case we will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words (one word which is a group of vowels)
These 6 words can be arranged in 6P6 ways
⇒6P6 = 6!/1! = 6! = 720 WAYS
Also, three vowels in a group may be arranged in 3! ways
⇒3! = 6 ways
Therefore, required number of words = (720*6) = 4320.