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[HELP] Probability question (1 Viewer)
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faisalabdul16
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Okay so you take two cases.
One where 2 boys sit next to each other and the other when all 3 sit next to each other.
Case 1)
From the three boys you choose 2. So 3C2. Then you fix one of the boys positions anywhere. The second boy has 2 spots to choose and sit next to the other guy. Either his left hand side or right. So far, 3C2 x 2. Now two boys are together, the third must not sit beside any of the boys and therefore he only has 4 places to choose from. Then, because there's no restrictions on girls, they can be arranged in 5! ways.
So for two boys next to each other, number of ways = 3C2 x 2 x 4 x 5! = 2880
Case 2) three boys together
Let boys be ABC AND GIRLS be dashes
If you place seats in a line it'll be ABC-----
Bold parts = the 8 seats. The three boys sitting together can be arranged in 3! ways. Then the rest of the girls can be arranged in 5! ways because it's a circle. So total ways of 3 boys together = 3! X 5! = 720
Total number of ways = 3600
Probability = 3600/7! (7! Because without restrictions that's total arrangements)
= 3600/5040
=5/7
One where 2 boys sit next to each other and the other when all 3 sit next to each other.
Case 1)
From the three boys you choose 2. So 3C2. Then you fix one of the boys positions anywhere. The second boy has 2 spots to choose and sit next to the other guy. Either his left hand side or right. So far, 3C2 x 2. Now two boys are together, the third must not sit beside any of the boys and therefore he only has 4 places to choose from. Then, because there's no restrictions on girls, they can be arranged in 5! ways.
So for two boys next to each other, number of ways = 3C2 x 2 x 4 x 5! = 2880
Case 2) three boys together
Let boys be ABC AND GIRLS be dashes
If you place seats in a line it'll be ABC-----
Bold parts = the 8 seats. The three boys sitting together can be arranged in 3! ways. Then the rest of the girls can be arranged in 5! ways because it's a circle. So total ways of 3 boys together = 3! X 5! = 720
Total number of ways = 3600
Probability = 3600/7! (7! Because without restrictions that's total arrangements)
= 3600/5040
=5/7
Another way - don't treat the boys (or the girls) as distinct, so there is no need to choose 2 from 3.
We are forming a "circular word" from the letters GGGGG BBB.
Case I - all the boys together - only 1 possibility
Case 2 - exactly two boys together - place GBBG - then there are 4 places to place the 4th boy, so 4 arrangements.
Case 3 - exactly one girl between two of the boys - place GBGBG - then there are 2 possibilities for the remaining 2 seats
It is impossible to arrange them with at least 2 girls between every pair of boys, so no more cases.
Answer: (1+4)/(1+4+2) = 5/7
We are forming a "circular word" from the letters GGGGG BBB.
Case I - all the boys together - only 1 possibility
Case 2 - exactly two boys together - place GBBG - then there are 4 places to place the 4th boy, so 4 arrangements.
Case 3 - exactly one girl between two of the boys - place GBGBG - then there are 2 possibilities for the remaining 2 seats
It is impossible to arrange them with at least 2 girls between every pair of boys, so no more cases.
Answer: (1+4)/(1+4+2) = 5/7
Just realised this is most likely BS. The different 'words' are not equally likely.
Carrotsticks
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The 5!3! is for all the boys together.
Now we split the boys into 2 groups, one with 2 and another with just 1. These 2 groups may not be next to each other to ensure that we are considering purely the case when it's 2 boys next to each other.
The girls can be arranged in a circular fashion in 4! ways.
There are 5 positions where we may place the 2 'groups' of boys such that exactly 2 are next to each other, so 5C2.
Permute the boys = 3!
Swap the groups of boys = 2!