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!

## 91 Comments on “[Aptitude] Permutation Combination made easy without Formulas”

super work boss thanks am little weak in aptitude, am not using more words to boost you, just your work is simply superb

thanks a lot sir ji

Its just superb. no other words are coming to describe how beautifully this topic has been explained.

We all readers owe success (if there will be any)to you. Keep doing the good work.

thanks Mrunal

Regards

Abhishek

great work Sir……thank u for your posts

and also there is printer's devil in the ready made formula section for combination i think it should be nCr instead of nPr

thank u very much for ur guidance :)

Mrunal Sir,

I cant thank you enough for all the invaluable help you give!!! So easy and So simple, your style of teaching is JUST FABULOUS!!

Great work thanks for making P&C so easy for us

Regards

Dinesh

Thank you

knew how to do the questions…..but these tips have surely helped more in concept clarity

Dear Mrunal G !

Really today i m compelled by my heart to say thanks to u…

Our education system needs innovative approach to impart lessons.

Learning can be such a fun ! simply superb…

Thanks again

Dear Mrunal

The work you have been doing is amazing!!! I have been following your site for past 6 months and the simplicity with which topics are explained by you..can beat any book or coaching institution! Just wanted to say…keep doing the good work, even when you join a B-school (assuming from your post that you might be joining one soon :))

Manish

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A. 210 B. 1050

C. 25200 D. 21400

sir can u plzzz tell me watz d answ with explanatn plzzz

3 CONSONANTS CAN BE SELECTED FROM 7 SO = 7C3

2 VOWELS CAN BE SELECTED FROM 4 IN = 2C4

NOW THESE ALL WORDS CAN BE PLACED IN 5P5

SO NOW 7C4 X 4C2 X 5P5= 25200

7C3 X 4C2 = 210

The answer is given in following article

[Aptitude] PnC: How many Ways to arrange letters

http://mrunal-exam.blogspot.com/2012/03/aptitude-pnc-how-many-ways-to-arrange.html

mrunal sir ,i always look forward eagerly to reading your articles.your economics chapters are invaluable as the language used is so interesting…by the way i myself got 95 percentile in cat this year (final yr of btech) and i'm in a dilemma over ias and cat next year

thanks sir,,,,,,,,,

i highly and heartly appreciate ur effort.

i m an iit delhi student in maths steam bt never seen easy presentation like this.

plz keep ur effort continued,,,,,,,,,

I have always struggled to learn these formulas…. Thanks Mrunal to male life easy!!.. Cheers!!

U made really easy. my concept got clear. thanks a lot

CSAT: General Studies 2012 (Paper 2) by Arun Sharma and Abhijit Guha (Tatamacgrawhill Publication)

Being a software Engineer and having finished half of this book, I can say this is the worst book I have ever read in whole of my life.. Its full of mistakes (not printing errors) specially solutions to given problems

E.g.

General Mental Ability page 60 question 9-15.

since it has 7 questions I wasted couple of mins in solving the problem. I did not find my answers in given options. Then I looked at solution at page 64. The author has done very stupid mistake.. and surprisingly he has given answer options based on his solution.

At this critical time (25 days left for the D day) wasting 10-15 minutes on such question really suck..

Thanks

Sanjeev

Mrunalji,

We can solve hand shakes and other problems from the definition of the combination / permutations(if reqd.) .

For eg. what i did –> Since we require at least 2 person for 1 hand shake ( also we cannot take greater than 2 person, because that would not look like a hand shake) so selecting 2 person out of 6 = 6C2 = 15 no. of ways hand shakes can be done.

So the process of over-counting is eliminated.

And if there is any discrepancy in my approach Mrunalji, do let me know.

Mrunalji,

We can solve hand shakes and other problems from the definition of the combination / permutations(if reqd.) .

For eg. what i did –> Since we require at least 2 person for 1 hand shake ( also we cannot take greater than 2 person, because that would not look like a hand shake) so selecting 2 person out of 6 = 6C2 = 15 no. of ways hand shakes can be done.

So the process of over-counting is eliminated.

And if there is any discrepancy in my approach Mrunalji, do let me know.

Great help from you sir!!!

sir ill u plz suggest me which book should i refer for gate exams?

thank u sir……………

thank u sir …..for this helpful article…

Book on General Studies 2012 (Paper 2) by Arun Sharma and Abhijit Guha not available on flip kart. It shows out of print.

dear mrunal ji.. in case 3 image ..i guess by typing mistake you had wrote 6men and 6 chair instead of 5 men and 5 chair

thanks for p&c made so easy !

Thank you Sir! you explained it much better than books as they only give formulas and don’t clarify the concepts behind it. :)

very useful.permutations was a big problem earlier..now im very much clear about the concept..thank u very much sir …u simplify things in a simple way .i think thats something what students require.

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

sir can u tell me how to solve this one without formula.

3 cases to be considered:

1. all 5 women; number of selection: 7c5= 21

2. 4 men, 1 women; number of selections: 7c4x6c1= 210

3. 3 men, 2 women; number of selections: 7c3x6c2= 525

total = 525+210+21= 756

Dear Mrunal HOW TO prepare ifs I have tried many time ask many question from you but I send my problem and how I see your reply answer please help me

Dear sir,

thanks a lot for your great exaplanation.