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At a school in Utah there is a bench that can fit 2 people. We took an exam and concluded that it is impossible to copy from someone that is not sitting in the same bench as you. So, the teacher gave assigned sits which were at random. The class consists of twenty students and ten benches. We have that there are 12 A's and 8 B's. Now we have 8 A's sitting at the same bench and 4 B's sitting at the same bench. What is the probability that this will occur?

I asked a question similar to this yesterday but my professor emailed me back today saying that question was impossible to do and he revised the question to this. How will I be able to do this?

With the help of Mathguy and Andre I know that:

I must numerate the seats so the first bench will sit 1 and 2, the second bench sits 3 and 4, third bench sits 5 and 6. (An analogy of this problem will be if I had 20 cards with the number 1 up to 20, where each number corresponds to a student. Thus I know that 12 cards has an A and 8 cards has a B). Then I must calculate that 8 A's are sitting at the same bench, then I must calculate/count the total number of ways to place these 20 different cards on 20 different benches. Then I must calculate the number of possibilities to place them so that I have 8 A's together and B's together. Then, if I denote j to be the number of A's on even seats, where each bench has exactly one even seat. Then I can count the number of possibiites that lead to 8 A's together and then I will sum over the different possible j's.

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Sounds like your teacher is making you compute to which benches he should assign you. –  Raskolnikov Nov 13 '12 at 23:44
    
@Raskolnikov haha. –  Q.matin Nov 13 '12 at 23:49
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1 Answer

up vote 1 down vote accepted

There are $20!$ permutations of people on seats. To seat all the $B$'s together, you choose the four benches they sit on in ${10 \choose 4}=210$ ways, then can seat the $B$'s on them in $8!$ ways and can seat the $A$'s on the others in $12!$ ways. $\frac {210(8!)(12!)}{20!}= \frac7{4199}$

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