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13 Sep 2022, 08:48
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D
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Six kids will be lined up from left to right. Boys may not be at either end (the first and last spot). In how many ways can the kids be lined up?
(1) There are 4 girls
(2) There are 12 different ways to fill the first and last spot
(1) There are 4 girls
(2) There are 12 different ways to fill the first and last spot
13 Sep 2022, 11:28
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Six kids will be lined up from left to right. Boys may not be at either end (the first and last spot). In how many ways can the kids be lined up?
(1) There are 4 girls
=> 4 Girls and 2 Boys
First considering the constraints i.e., spots occupied by boys.
First boy can occupy any of 4 spots and second boy can occupy any of the 3 spots. Total number of ways in which Boys can be arranged = 4*3 = 12
4 girls can be arranged in remaining 4 spots in 4! ways.
Therefore total number of ways in which Kids can be lined up = 4!*12
Statement 1 alone is sufficient.
(2) There are 12 different ways to fill the first and last spot
=> Let there be G girls. The first and last spot can be occupied only by two of these Girls.
For the first spot we have G different girls. Then, for the last spot we have (G-1) different girls.
Total number of ways in which the first and last spots can be filled = G(G-1)
=> G(G-1) = 4*3 = 12
=> G=4
This information is same as Statement 1. Hence sufficient.
Answer: D
(1) There are 4 girls
=> 4 Girls and 2 Boys
First considering the constraints i.e., spots occupied by boys.
First boy can occupy any of 4 spots and second boy can occupy any of the 3 spots. Total number of ways in which Boys can be arranged = 4*3 = 12
4 girls can be arranged in remaining 4 spots in 4! ways.
Therefore total number of ways in which Kids can be lined up = 4!*12
Statement 1 alone is sufficient.
(2) There are 12 different ways to fill the first and last spot
=> Let there be G girls. The first and last spot can be occupied only by two of these Girls.
For the first spot we have G different girls. Then, for the last spot we have (G-1) different girls.
Total number of ways in which the first and last spots can be filled = G(G-1)
=> G(G-1) = 4*3 = 12
=> G=4
This information is same as Statement 1. Hence sufficient.
Answer: D
General Discussion
25 Nov 2025, 08:07
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Hi Bunuel, I was facing confusion with the wording 'may not'. This is not confirmative that boys will not be there at either ends.
