If the following questions make you uncomfortable, then read this article.

- In how many ways can 5 member committee be formed out of 20 students? Or
- How many ways can the word “UPSC” be arranged so that “U” is always second from left.
- In a party of 20 guest, if every member shakes hands with other, how many total handshakes are done?

First the basics

# FUNDAMENTAL COUNTING PRINCIPLE

- There are 3 trains from Mumbai to Ahmedabad and 7 trains from Ahmedabad to Kutch. In how many ways can Jethalal reach Kutch?
- From Mumbai to A’bad, Jethalal can go in 3 ways. (Because there are three train, he can pick anyone)
- Similarly From A’bad to Kutch, he can go in 7 ways.
- Total ways: simple multiplication =3 ways x 7 ways =21 ways Jethalal can reach Kutch.
- This is fundamental counting principle.

Let’s extend this further to sitting arrangement problems:

# PERMUTATION (PLACE, ARRANGEMENT)

6 Men of Gokuldham society go to Abdul’s sodashop. And there are 6 chairs. Yes I’m talking about our beloved *Jethalal, Bhide Master, Sodhi, Dr.Haathi, Mehta saab and Aiyyar.*

In how many ways can they be seated? Or How many way can you arrange these 6 gentlemen into 6 chairs?

## CASE 1: SIX MEN AND 6 CHAIRS

Let’s do one chair at a time.

For the first chair, you’ve 6 candidates:

Just pick one man and make him sit.

How many ways can you do it? Ans. 6 ways, bcoz you’ve 6 men and pick one. (Just like the train)

Now second chair: you’ve to pick one guy from the remaining 5 members. So 5 ways.

Third chair: 4 men remaining and you’ve to pick one: again 4 ways..

…

For the 6th chair only one man remains. So you can pick in 1 way only.

Now Just extending the fundamental counting principle

So total number of ways in which men of Gokuldham society can sit in chairs

= 6 x 5 x 4 x 3 x 2 x 1

=6! (six factorial ways)

## CASE 2: SIX MEN AND 3 CHAIRS?

How many arrangements possible?

= 6 x 5 x 4 x ..OK STOP! Because there are only 3 chairs.

Why?

First chair= 6 men pick one=6 ways.

Second chair = 5 men pick one=5 ways

Third chair=4 men pick one= 4 ways

That’s all.

So number of ways

=6x5x4

=120 ways.

CASE 3: ONE CHAIR ALWAYS OCCUPIED

In the 6 men 6 chairs problem, Dr.Haathi insists that he’ll sit in the number #1 chair only. Then How many arrangements are possible?

First chair= 1 man (Haathi) and you’ve to pick one= 1 way only!

Second chair=5 men remain, pick one =5 ways

third chair=4 men remain, pick one = 4 ways and so on…

…

So here we’ve

=1x5x4x3x2x1

=5! Ways.

=120 ways.

This is permutation. Here ‘order / ranking ’ matters. i.e. who sits in the first chair, who sits in the second chair etc.

The same question can appear under different wordings example

1. How many 4 lettered words can be formed out of “UPSC”, without repetition?

Answer: Same logic. Consider U, P, S, C are four gentlemen and they’ve to be arranged in 4 seats.

2. How many 4 letter words can be formed out of “UPSC” so that “U” always occupies the second position from left?

Answer: Same Dr.Haathi logic.

Second Topic is

COMBINATION (CHOICE)

CASE 1: COMMITTEE FORMATION

The one and only Secretary of Gokuldham society, Master Bhide decides to form a 3 member-Committee for arrangement of Holi-festival. Out of the 5 members (Jetha, Sodhi, Mehta, Popat and Aiyyar) how many ways can he do this?

Ans.

Here order or ranking doesn’t matter.

Because in case of chair sitting: we can say yes Mr.Sodhi is in first chair, Mehta saab in Second chair and so on…

But in case of Committee: it is only “IN or OUT” i.e. Yes Sodhi is in the Committee, No Mehta saab is not in the Committee.

So order doesn’t matter. Only selection matters.

Lets proceed

How many ways can you pick up the first member? = 5 ways.

Second member? = 4 men remain, 4 ways

Third member? = 3 ways.

So total ways= 5 x 4 x 3= 60.

But wait, there is over counting.

A committee made up of Mehta, Sodhi and Aiyyar (MSA) is same as a Committee made up of Aiyyer, Mehta and Sodhi. (AMS) (because order doesn’t matter, only selection matters: are you “In or Out?”).

But in this answer 60, we’ve over counted such ‘orders’.

That’s why we need to divide the answer

Suppose Mehta, Sodhi and Aiyyar are “in” the committee.

Suppose they’ve to sit in three chairs. How many ways can you do it?

Just like permutation in first example

3 x 2 x 1=6.

That’s the overcounting : 6.

Our answer is

=(5*4*3)/(3*2*1)=10 ways.

CASE 2: HANDSHAKES

Handshakes is another type of combination problem because

Aiyyar shakes hand with Sodhi or Sodhi shakes hand with Aiyyer= both incidents are one and same. “order” doesn’t matter!

So How many ways can 6 men of Gokuldham society shake hands with each other?

To shake hands, you need two men.

How many ways can you pick up first man? = 6 ways.

How many ways can you pick second man? = 5 men remaining so 5 ways.

Hence answer is 6 x 5=30

But wait, over counting!!

Suppose Sodhi and Aiyyer (SA) are selected for handshaking: =two men.

How many ways can you make two men sit in two chairs?

= 2 x 1 =2 ways. (SA and AS)

That’s the over counting.

So we’ve to divide this over counting: to get the correct answer.

Total Handshakes=(6*5)/(2*1)=15

The Formulas

The books give readymade formulas:

But essentially they’re derived from above concepts of fundamental counting principle.

Permutation =place them = order matters in placement

Combination= choice = only “Yes or No” or “In or Out”

## What now?

- Open your books right now, solve the sums using both the method shown above as well as the readymade formulas, then you’ll get a good command over these topics.
**Remember,**You can never learn the aptitude by reading the ‘sums’.- You must to get your hands dirty, and learn from the mistakes in calculation, so start solving the sums on paper by yourself.
*To be continued……*

What now?

Well I hope I was able to explain the topic-concept to you.

But it is not sufficient to crack the exam! Still You might end up making silly mistakes in calculation or get stuck in some question during the actual exam. So You must practice as many sums as you can at home, to get a firm command over the topic. That’s why Get the books, and start solving sums!

## So far 90 Comments posted

Thanks murnal sir.

this site has always been useful for me…to make my maths more stronger by gud examples………..special thnx to mrunal sir.please keep on doiing dis………nd another thng dat here d way of explanation is awesommmeeeeeeeee…………..

thank u sir!!!!!

Sir the article above it is not shown..please help

Example 5.1

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

Solution:

Let the beds be numbered 1 to 7.

Case 1: Suppose Anju is allotted bed number 1.

Then, Parvin cannot be allotted bed number 2.

So Parvin can be allotted a bed in 5 ways.

After allotting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5*5!=600 ways.

Case 2: Anju is allotted bed number 7.

Then, Parvin cannot be allotted bed number 6

As in Case 1, the beds can be allotted in 600 ways.

Case 3: Anju is allotted one of the beds numbered 2,3,4,5 or 6

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed.

For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted 4*5!=480 ways.

=> The beds can be allotted in:

2*600+5*480=1200+2400= 3600 ways

sir is question me ye last step kyun kiya gaya h. plz samjha dijiye

the best way to do this type of questions is…

First assume there are 7 people who has to be arranged with no conditions..

that will be 7! = 7*6*5*4*3*2*1 = 5040 ways

Now the problem comes…

Lets make Anju and Parvin a bundle and allocate the beds…

in 6! ways = 6*5*4*3*2*1 = 720 ways

But Anju and Parvin within themselves can be arranged 2 ways. i.e., 2!= 2

Now total ways keeping Parvin and Anju together is 720*2 = 1440.

But we want the other way round… Total ways keeping Parvin and Anju separate…

That will be 7! – (6!*2!)= 5040- (720*2) = 3600 ways… 🙂

really best method and quicker. thanx

sir,thank you for providing guidance in every topics concerned.

Can you solve a question for me :

Numbers of ordered pairs of positive integers ( a, b, c, d ) such that LCM of a, b, c, d is 729

I studied this topic many times but never understood before……but sir you r too good….you make things so simple….may god give you a good gf..

Sir, I have been following your articles on almost every section & each one of them is well designed in a methodical manner.Please keep continuing the good work.

( ) + ( ) + ( ) + ( ) + ( ) = 30

USE 1,3,5,7,9,11,& 13 REPEAT RE USE EVERYTHING ALLOWED… ASKED IN upsc EXAM PLEASE SEND ANSWER

11+9+7+3+1=30

11+5+9+1+[(13+3)/(3+1)]=30

3 weeks+7 days+11hours+13 hours+1day=30 days

How many different four letter words can be formed (the words need not be meaningful) using the letters of the word “MEDITERRANEAN” such that the first letter is E and the last letter is R?

sol-

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options – the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of posssibilities = 56 + 3 = 59

my query is- how can we assume that question is asking about 2 cases-

1) diff letters at second and third place

2) same letters at second and third place.

here if i take first case as no repetition case than the answer will be 56 but if i take second case as- repetition, than i’ll not get the answer as 3 MENTIONED IN ABOVE SOLUTION.

Is same letter case is different than repetition (am assuming)?

kindly help with this….

Kindly tell me…you have any own written book that I can purchase..because your tricks are awesome…..kindly reply to my id [email protected]

sir the questions from simplification with approximation for 5 marks are some time goes wrong …. any suggestion plzz. and do u have any book …. from where I can get your tricks

i cant access some articles about permutation and combination.. there is some error while open this articles pls help me

2sin(πx/2)=x2 +1/x2

Then what is value of the (x – 1/x)

Are there any short tricks for surface areas and volumes?

There are 6 person and 6 chair. The first person can not be seated in first chair, and the second person in second chair and so on. How many different ways of 6 person occupying 6 chairs?

-3/2

It’s n(n+1)/2… not n(n-1)/2

What is your Message? Search before asking questions & confine discussions to exams related matter only.

sir,in the above theory it’s written that in combination ,order doesn’t matter and in permutation it does..i have read (in books) just the opposite.

thnqqq sir……..pls keep continuing the good work

thnqq sir

what happened to images in your article, same problem in mixture and allegation article.

Thank you Mrunal Sir…. Ur site is very useful for whoever is preparing for competitive exams…. Images are not displayed…. please rectify the problem….

I have never seen before like this Thank you very much Sir !!!!

Hello sir,

I am from Nepal

Please teach me how to do this:

If 50 different jewels can be set to form a Necklace then the number of ways is?

a) 1/2 × 50! b) 1/2 × 49! c) 49! d) None of these.

Thank you 🙂

I have problem of permutation nd combination. Please help me

thq 4r the short cut sir

formation of 4 member committee,,,from 3 of production department,,4 of purchase department,,2 of sales department,,,and 1 CA,,,,,the number of chances there must be atleast one from each category