Cialis prezzo svizzera 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)
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DONOT Proceed further, without reading The First article on basic concept of Permutation and Combination
- Always Sit together
- Never Sit together
- Demo Questions: Permutation sitting together
- Always Select
- Never Select
- Demo Question: Selection
here CASE: ALWAYS SIT TOGETHER
how to buy cialis in canada 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.
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
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.
Que. Same 3 couples (=6 persons), 6 chairs. How many ways you can arrange them so that Dayaben and Jethalal (DJ) never sit together.
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.
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.
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
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
2 must be selected (DJ)
How many ways can you select Dayaben and Jethalal? : One way.
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.
=(Daya Jetha selection ways) x (2 out of 4 selection)
=1 x 6
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
Q. Select 4 out of 6 given characters for Kutch-episode, such that Dayaben and Jethalal are never selected in the crew?
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.
- 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)