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07 Apr 2022, 02:45
Kudos
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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:
55%
(hard)
Question Stats:
62% (02:07) correct
38%
(02:13)
wrong
based on 145
sessions
History
Date
Time
Result
Not Attempted Yet
Father, mother and 6 children stand in a ring. In how many ways can they be arranged if father and mother are not to stand together? (2 seating arrangements are considered different only when the positions of the people are different relative to each other.)
A. 1440
B. 1800
C. 2400
D. 3600
E. 5040
A. 1440
B. 1800
C. 2400
D. 3600
E. 5040
19 Apr 2022, 09:52
Kudos
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Bunuel
If there are \(n\) people, then the number of ways to arrange them in a ring is \((n-1)!\)
So total ways to arrange the \(8\) family members in a ring is \((8-1)! = 7!\) without any restrictions.
Lets club the Father & Mother together, and consider them \(1\) member .
Now we have a total of \(7 \) members, and number of ways to arrange them in a ring is \((7-1)! =6!\)
Additionally Father could be to the left and mother could be to the right or vice versa.
So among themselves Father , Mother can be arranged in \(2!\) ways.
So total ways to arrange family keeping Father and Mother together is \(6!*2!\)
Unrestricted total ways - Total ways Father and Mother together = Total ways father Mother are NOT together.
\(7!- 6!2! = \)
\(7 *6! -6!2! =\)
\(6!( 7-2) = 6!5= 3600\)
Ans D
08 Sep 2024, 02:10
Kudos
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