Six students - 3 boys and 3 girls - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
A) 12
B) 36
C) 72
D) 240
E) 720
The OA si C.
I'm confused with this PS question. Experts, any suggestion about how to solve it? Thanks in advance.
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Six students - 3 boys and 3 girls - are to sit side by...
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BTGmoderatorLU
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The boys and girls must ALTERNATE.LUANDATO wrote:Six students - 3 boys and 3 girls - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
A) 12
B) 36
C) 72
D) 240
E) 720
Case 1: BGBGBG
Within the 3 red positions, the number of ways to arrange the 3 boys = 3! = 6.
Within the 3 blue positions, the number of ways to arrange the 3 girls = 3! = 6.
To combine the 6 options for the boys with the 6 options for the girls, we multiply:
6*6 = 36.
Case 2: GBGBGB
Case 2 will yield the same number of ways as Case 1:
36.
Total ways = Case 1 + Case 2 = 36+36 = 72.
The correct answer is C.
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Since no two boys and no two girls can sit next to one another we can have the following arrangements:LUANDATO wrote:Six students - 3 boys and 3 girls - are to sit side by side for a makeup exam. How many ways could they arrange themselves given that no two boys and no two girls can sit next to one another?
A) 12
B) 36
C) 72
D) 240
E) 720
B-G-B-G-B-G or G-B-G-B-G-B
We see that either arrangement can be selected in the following ways:
We have 3 options for the first boy, then 3 options for the first girl, then 2 options for a boy, 2 options for a girl, and then finally 1 option for each boy and girl.
This can be done in 3 x 3 x 2 x 2 x 1 x 1 = 36 ways.
Since the second arrangement can also be done in 36 ways, there are 36 + 36 = 72 possible ways of making the arrangement of no two boys and no two girls sitting next to one another.
Answer: C
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