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19 Feb 2017, 10:50
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:20) correct
27%
(01:47)
wrong
based on 357
sessions
History
Date
Time
Result
Not Attempted Yet
John and four friends go to a Lakers game. In how many ways can they be seated in five consecutive seats, if John has to sit between any two of his friends?
(A) 144
(B) 120
(C) 96
(D) 72
(E) 48
(A) 144
(B) 120
(C) 96
(D) 72
(E) 48
19 Feb 2017, 11:10
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5! is the number of ways to place the five people => 120
There is two impossible case when John sit in the left and in the right => Therefore the number of possible case is 3
120 * 3/5 (the probability of possible case)= 72 => Answer D
There is two impossible case when John sit in the left and in the right => Therefore the number of possible case is 3
120 * 3/5 (the probability of possible case)= 72 => Answer D
19 Feb 2017, 12:19
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vikasp99
Hey,
PFB the solution.
Let us first make 5 spaces representing the 5 seats.
___ ___ ___ ___ ___
As per the given condition, John has to sit between any two of his friends, that mean John cannot sit at the first and the last seat.
Therefore, the total number of ways in which John can sit is 3.
__ _J_ __ __ __ OR __ __ _J_ __ __ OR __ __ __ _J_ __
Now we are left with 4 friends who can occupy any of the four seats without any restriction in \(^4P_4\) ways \(= 4! = 24\)
Thus the total number of ways in which these 5 friends can be seated
- = Number of ways in which John can sit AND Number of ways in which the other 4 can sit
\(= 3 * 24\)
\(= 72\)
Hence, the correct answer is Option D.
Thanks,
Saquib
Quant Expert
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