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10 Jun 2025, 23:12
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D
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28% (02:02) correct
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A team consists of 8 males and some females. The team decides to create a lineup with all its members so that no two males are paired together, and no two females are paired together. If the number of females in the team is not less than the number of males, how many such lineups are possible?
(A) 8! x 8!
(B) 8! x 9!
(C) 8! x 8! x 2
(D) 8! x 8! x 11
(E) 8! x 8! x 10
(A) 8! x 8!
(B) 8! x 9!
(C) 8! x 8! x 2
(D) 8! x 8! x 11
(E) 8! x 8! x 10
11 Jun 2025, 00:48
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Condition: No 2 males are together AND no 2 females are together.
Think - When will this happen? Males and Females must be in alternate positions. Else, at least 2 males (or females) will end up together.
Visualizing the possible scenarios -
_M_M_M_M_M_M_M_M_
The spaces highlighted above is where we can have females. We can have at most one female in any of these spaces.
Given that number of females >= number of males, here are the possible scenarios ->
(1) F M F M F M F M F M F M F M F M _ (8 males and 8 females)
Number of different lineups possible = 8! (ways of placing the males) x 8! (ways of placing the females) = 8! x 8!
(2) _ M F M F M F M F M F M F M F M F (8 males and 8 females)
Number of different lineups possible = 8! (ways of placing the males) x 8! (ways of placing the females) = 8! x 8!
(3) F M F M F M F M F M F M F M F M F (8 males and 9 females)
Number of different lineups possible = 8! (ways of placing the males) x 9! (ways of placing the females) = 8! x 9!
Final answer = (1) + (2) + (3) = 8! x 8! (1 + 1 + 9) = 8! x 8! x 11. Choice D.
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Harsha
Think - When will this happen? Males and Females must be in alternate positions. Else, at least 2 males (or females) will end up together.
Visualizing the possible scenarios -
_M_M_M_M_M_M_M_M_
The spaces highlighted above is where we can have females. We can have at most one female in any of these spaces.
Given that number of females >= number of males, here are the possible scenarios ->
(1) F M F M F M F M F M F M F M F M _ (8 males and 8 females)
Number of different lineups possible = 8! (ways of placing the males) x 8! (ways of placing the females) = 8! x 8!
(2) _ M F M F M F M F M F M F M F M F (8 males and 8 females)
Number of different lineups possible = 8! (ways of placing the males) x 8! (ways of placing the females) = 8! x 8!
(3) F M F M F M F M F M F M F M F M F (8 males and 9 females)
Number of different lineups possible = 8! (ways of placing the males) x 9! (ways of placing the females) = 8! x 9!
Final answer = (1) + (2) + (3) = 8! x 8! (1 + 1 + 9) = 8! x 8! x 11. Choice D.
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Harsha
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