There are x children and y chairs arranged in a circle in a room where

Question

There are x children and y chairs arranged in a circle in a room where x and y are prime numbers. In how many ways can the x children be seated in the y chairs (assuming that each chair can seat exactly one child)?

(1) x + y = 12
(2) There are more chairs than children

(1) x + y = 12
(2) There are more chairs than children

KarishmaB · Accepted Answer

Responding to a pm:

Stmnt 1:

If x = 7 and y = 5, you first select 5 of the 7 children in 7C5 ways = 21 ways.
5 children can be seated in 5 chairs in a circle in 4! ways.
There are 21 * 4! = 504 ways

If x = 5 and y = 7, you can sit the first child in 1 way. Then there are 6 distinct chairs and 4 children. You can seat them in 6*5*4*3 = 360 ways

We get different answers. Not sufficient.

Stmnt 2:
Not sufficient.

Both together, we know that we have 7 chairs and 5 children. 
Answer (C)

MarmikUpadhyay · Answer

Given: x children and y chairs arranged in a circle, x and y are prime numbers.
In how many ways can x children be seated in y chairs.

Statement 1:
x + y = 12
There is only one pair of prime numbers whose sum is 12. It is 5 and 7.
Case 1:
x = 5, y = 7
Ways in which 5 children can be seated in 7 chairs, arranged in a circular manner =  7 P 5 /7 (Chairs are in a circle, so one case will be counted 7 times, therefore dividing by 7)
=>  7 P 5 /7 =  \frac{7!}{2! * 7}  = 360 ways.
Case 2:
x = 7, y = 5
Ways in which 7 children can be seated in 5 chairs, arranged in a circular manner =  7 C 5  * 5!/5 (Chairs are in a circle, so one case will be counted 5 times, therefore dividing by 5)
 7 C 5  * 5!/5 =  \frac{7! * 5!}{5! * 2! * 5}  = 504 ways.
Statement 1 is Not Sufficient.

Statement 2:
y > x
There is no much information on number of children and chairs.
Statement 2 is also Not Sufficient.

Statement 1 and Statement 2 combined:
x + y = 12 and y > x
This is the same as Case 1 in Statement 1.
Ways in which 5 children can be seated in 7 chairs, arranged in a circular manner = 360 ways.

So, correct answer is option  C .

tyson111 · Answer

A)x+y=12
Not sufficient

X=5 or 7
Y=7 or 5

B)y>x
Not sufficient

Together
Y>X

X=5
Y=7

Posted from my mobile device

sumitkrocks · Answer

Since X and Y are prime numbers , the two values MUST be 5 and 7.

so there may be two cases 
Case 1:
If
X = 5
Y = 7

This means that there are 2 empty chairs, so we'd first have to choose 5 chairs (which is 7C5 = 21 different sets of 5 chairs) and then arrange the 5 children in those 5 chairs (which is 5! = 120 different arrangements. Under these circumstances, there are (21)(120) different arrangements.

Case 2
If
X = 7
Y = 5

This means that there are 2 more children than seats, so 2 of the children won't be seated. The number of arrangements would be (7)(6)(5)(4)(3)=(21)(120) same as in case 1

So, either option yields the SAME RESULT.

Statement 1 is sufficient

Statement 2:
Clearly not sufficient, so Option A is correct in my opinion

MarmikUpadhyay · Answer

As chairs are in circular arrangement it is (n - 1)!, so it will be 4! or  \frac{5!}{5}  and not 5!, for highlighted in red.

And for case 2, the total arrangements will be divided by 7, because 7 arrangements will be recounted every time in circular arrangement.

KetakiUpadhye · Answer

Why have you used Permutation in one and combination in the other.

Posted from my mobile device

MarmikUpadhyay · Answer

For case 1, we are simply arranging 5 children in 7 chairs, so directly Permutation. But for case 2, we will have to select the 5 children from 7 children first, so combination and then arranging them in 5!/5 ways. 
Hope it helps!

KetakiUpadhye · Answer

Thank you very much Marmik.

Posted from my mobile device

Rebaz · Answer

KarishmaB

Can you please elaborate on:  If x = 5 and y = 7, you can sit the first child in 1 way. Then there are 6 distinct chairs and 4 children. You can seat them in 6*5*4*3 = 360 ways ?

What do you mean with  you can sit the first child in 1 way .

I donot get your approach.

Thanks in advance!

KarishmaB · Answer

We are talking about seating the children in a circle. In a circle, all seats are considered identical (with respect to each other). Think about it - say you see a round table with 7 seats around it. Are all seats equivalent? Sure. You can sit on any; they are all the same with respect to each other. Hence the first person takes a seat around a circular table in 1 way only. But once you do, now all seats are different. One seat is to the immediate right of the child and another to the immediate left. One seat is second to the right of the child and another to the left etc. So now you have 6 distinct chairs and 4 children to be seated on them. So you select 4 chairs in 6C4 ways and arrange them in 4! ways (Combinations). This is equivalent to 6*5*4*3 method (Basic counting principle) Both Basic Counting Principle and Combinations are discussed in these two videos: Video on Basic Counting Principle: https://youtu.be/LFnLKx06EMUVideo on Combinations: https://youtu.be/tUPJhcUxllQ

Rebaz · Answer

KarishmaB

Thank you Karishma for your  Superb  explanation!

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