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12 May 2021, 08:45
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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:
35%
(medium)
Question Stats:
69% (01:33) correct
31%
(01:46)
wrong
based on 115
sessions
History
Date
Time
Result
Not Attempted Yet
David and Rachel are getting married. The extended family wants to have its picture taken, but David's father is unwilling to stand next to Rachel's mother. How many options does the photographer have to arrange the 10 family members in a row for the picture?
A. 9!
B. 9×8!
C. 8×9!
D. 10!/2!
E. 10!
A. 9!
B. 9×8!
C. 8×9!
D. 10!/2!
E. 10!
15 May 2021, 07:00
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Bunuel
In how many ways David's father and Rachel's mother can stand together? Consider them as one unit: {F, M}. So, we'd have 9 units {F, M}, 1, 2, 3, 4, 5, 6, 7, 8, which can be arranged in 9! ways. David's father and Rachel's mother within their unit can be arranged in 2! ways. Therefore, David's father and Rachel's mother can stand together in 9!*2! ways.
Total - Restriction =
= 10! - 9!*2! =
= 9!(10 - 2!) =
= 8*9!.
Answer: C.
General Discussion
12 May 2021, 08:54
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Bunuel
Total number of possible arrangements - Arrangement where David's father stands next to R's mother = Arrangement possible where David's father doesn't stand next to R's mother.
Total arrangements possible = 10!
Arrangement where D's father stands next to R's mother = 9! * 2
So, 10! - 9!*2
= 9!(10 - 2)
=9! * 8 (C), IMO
