# [Aptitude] PnC: arrange such that two people Always sitting together?

Q. How many ways can four children be made to stand in a line such that two of them A and B are always together? (-asked in UPSC Prelims 2010)

Confused? You won’t be, after reading this.
DONOT Proceed further, without reading The First article on basic concept of Permutation and Combination

CASE: ALWAYS SIT TOGETHER

Three couples are standing in the compound of Gokuldham society

• Jethalala + Dayaben
• Mehta saab + Anjali
• Aiyyer + Babita-ji How many ways can you arrange them in such a way that Jethalal and Dayaben are always sitting together?
First, ask yourself, is it a permutation problem or a combination problem?
Since sitting= order matters, this is a permutation problem.
Now let us consider Jethalal and Daya as “ONE Person”. And they’ll sit in “one chair”. Now you’ve the case of 5 Persons and 5 Chairs.
How many ways can use arrange them?
As we saw in the previous article
= 5 x 4 x 3 x 2 x 1
=5! Factorial Ways.
=120
But here ‘order’ matters, this is a permutation problem so
In the sitting order Daya-Jetha (DJ) and Jetha-Daya (JD) are two different arrangements.
So now time to split that “One person” Apart.
How many ways can use sit 2 persons (DJ) in 2 chairs?
2 x 1
=2 ways.
So, here we’ve
5 people sit in 5! Ways (=120)
2 people (Daya-Jetha) can sit in 2 ways.
According to fundamental counting
120 x 2 = 240 ways you can arrange this.

(Formula: 2P2 x 5P5); but we don’t need to use it.

CASE: DON’T SIT TOGETHER

Que. Same 3 couples (=6 persons), 6 chairs. How many ways you can arrange them so that Dayaben and Jethalal (DJ) never sit together.
Ans.
First how many ways can 6 people sit in 6 chairs?
=6! Factorial Ways. (=720) ; Solved in the first article.

How many ways can 6 people sit such that Dayaben and Jethalal (DJ) always sit together
=240 ways ; solved in previous case.

So
Total ways = Cases where DJ together + Cases where DJ not together
720=240 + DJ not together
DJ not together = 720-240=480 ways.
Answer: in 480 ways you can arrange the given 6 people such that Dayaben and Jethalal don’t sit together.
Formula: 6P6 minus [2P2 x 5P5]; but we don’t need to use it.

Demo Question:Sitting together

In CSAT Paper II (Aptitude), this can appear under different situations

• How many ways can four children be made to stand in a line such that two of them A and B are always together? (-already asked in UPSC Prelims 2010)
• How many ways can you arrange the word “Trial” where letters T and L are always together?
• How many ways can you arrange the word “Trial” where letters R and A are never together?

COMBINATION PROBLEM: ALWAYS SELECT, NEVER SELECT

CASE: ALWAYS SELECT

Director Asit Modi wants to shoot an episode in Kutch. Out of the 6 characters, how many ways can he select 4 character such that Dayaben and Jethalal (DJ) are always in the crew?

Here ‘order’ doesn’t matter, only selection matters. After all this is just another ‘Committee’
DJ and JD are one and same.
Now deconstruct the problem
6 people
Select 4.
2 must be selected (DJ)

How many ways can you select Dayaben and Jethalal? : One way.
Matter finish.
Now worry about the remaining 2 vacancies.
we’ve 4 characters left (Mehta-Anjali, Aiyyar-Babita)
And we’ve to select two of them.
Classical combination problem. Solve it according to previous article.

(4*3)/(2*1)=6 ways.

Fundamental Counting
=(Daya Jetha selection ways) x (2 out of 4 selection)
=1 x 6
=6 ways.
Final Answer : Asit Modi can select the characters in 6 ways such that Jethalal, Dayaben are always in the crew. In case you’re curious, here is the graphical representation of it CASE: “NEVER SELECT” in the crew

Q. Select 4 out of 6 given characters for Kutch-episode, such that Dayaben and Jethalal are never selected in the crew?

Ans.
Very easy. Drop DJ, how many left?
=4 Characters: Mehta-Anjali, Aiyyar-Babita
4 Characters, 4 vacancies. How many ways can you select? Just one way: you’ll have to select all of them! because order doesn’t matter, this is a combination problem!

PS. If you wish to solve it by formula: it is 4C4=1; because nCr=1 when n=r.

Demo Questions for CSAT paper II (Aptitude)

• A zoo is has to purchase 3 animals out 5 given animals: Zebra, Elephant, Tiger, Lion and Leopard. How many ways can they do it? (5c3)
• ^Such that Zebra and Elephant are always selected? (3C1)
• ^Such that Tiger is never selected? (4C3)

## 26 Comments on “[Aptitude] PnC: arrange such that two people Always sitting together?”

1. HEARTFELT thanks… i have never grasped permu and combo as clear as this time here ! Keep going sir, im waiting for all the aptitude topics. THIS is the place im going to do my Csat prep :):)

2. The First article on basic concept of Permutation and Combination is not available plz look at this!

1. sir how can solve this p&c problem

Q. In how many ways can 6 students and 4 teachers be arranged in a row so that NO TWO teachers are togehter?
a)604800
b)24680
c)25860
d)none of these

i was trying like ,,nmbr of ways they can arrange means 10!-4c2(two teacher toghtr)= they are not toghtr but ansr is wrong..pls tell me where m misleading…thanks.

1. the answer should be D None of these as it should be 35.

let the Students stand in a single row. now when 6 Stud stand they create 7 spaces. Like this ( Student=S Teacher = T )

_S_S_S_S_S_S_

Now the blank spaces created are 7 and now when u put 4 teachers in it so that no two are together u ll have 7 places to be filled by for.
for example one ofe them could be = TSTSTSTSSS
This order or arrangement is not important = Combination.

7C4 = 35 is the correct answer.
I dont know what options they have given , seeing hte options one can easily mark the answer.

2. _ S_S_S_S_S_S_
6 students can be arranged in 6! Ways.
4 teachers needs to fill 7 places where no teachers together.so these 7 places can be filled by 7P4 ways i.e 7*6*5*4 ways. Now as per fundamental counting principle 6! * (7*6*5*4) = 720 * 840 = 604800

2. Please tell the situation when atleast one has to be selected from amongst Dayaben or Jethalal in the committee !!

3. in mock questions, in the first que. how 5c3 comes? because in your first permutation article 2nd problem 3 chairs and out of 6 men you have to select 3 and answer is 6X5X4 and here in this sum 5c3? 5X4X3 is wrong then how?

1. you will divide by 3X2X1 also as it is a combination problem and order doesnt matter here . So it will be (5x4x3)/(3x2x1)= 10 WAYS. BUT M HAVING PROBLEM IN tiger not selected question , can anybody explain it.

4. There r 25 points on a plane of which 7 r collinear, how many straight lines can b formed,
Can u explaine mrunal sir coz I get difficulty in ths combination problem. Pls sir do rply

5. a bag contains 5 red 3 yellow 2 white balls. No. of ways in which one or more balls can be taken out from the bag is?
options: 66, 72, 71, 30.