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29 Mar 2021, 00:21
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
66% (01:21) correct
34%
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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
31 Mar 2021, 20:58
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Bunuel
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)
General Discussion
29 Mar 2021, 07:19
Kudos
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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 = 7P5/7 (Chairs are in a circle, so one case will be counted 7 times, therefore dividing by 7)
=> 7P5/7 = \(\frac{7!}{2! * 7}\) = 360 ways.
Case 2:
x = 7, y = 5
Ways in which 7 children can be seated in 5 chairs, arranged in a circular manner = 7C5 * 5!/5 (Chairs are in a circle, so one case will be counted 5 times, therefore dividing by 5)
7C5 * 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.
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 = 7P5/7 (Chairs are in a circle, so one case will be counted 7 times, therefore dividing by 7)
=> 7P5/7 = \(\frac{7!}{2! * 7}\) = 360 ways.
Case 2:
x = 7, y = 5
Ways in which 7 children can be seated in 5 chairs, arranged in a circular manner = 7C5 * 5!/5 (Chairs are in a circle, so one case will be counted 5 times, therefore dividing by 5)
7C5 * 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.
