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I recently gave an exam wherein this sum was given. i could not solve it.please solve it for me.

Q) There are 6 people in a line. "A" and "B" are two out of them. Find the probability that they are seperated by only one person.

Options are

  1. 8/15
  2. 3/15
  3. 4/15
  4. 2/15

P.S time limit is one minute for the sum to be solved

P.S I don't vividly remember the options. If your answer does not match the options please ignore them.

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So what is unclear to you in this question? – Ilya May 26 '12 at 16:09
@Ilya i could not solve it... give me a hint or solve it for me – Ashu May 26 '12 at 16:09
Hint it to consider all possible positions for persons and count in how many of them A and B are separated by 1 person. Then divide the latter by the number of all possible positions. – Ilya May 26 '12 at 16:11
@Ilya Oh i forgot to tell the sum has to be done in a minute – Ashu May 26 '12 at 16:13
up vote 2 down vote accepted

In how many ways can you choose two objects (A, B) out of 6? Now, if you denote by _ _ _ _ _ _

the six places in the queue, in how many ways can you choose two places separated only by one

other place? Multiply now this by two (as, apparently, we don't care whether A is behind B or

the other way around) and divide by the total you first calculated...

Added: Simon's comment below made me think (!) about the number of possible placings of a pair of persons in a queue with 6 persons in it: there are 30, not 15, such possibilities, as we can easily check if we place say $\,A\,$ in the first place and then $\,B\,$ can be placed in 5 different places in the queue (yes, it is not the same $ A$ in first place, $B$ in second than the other way around, unless some further info is given). Thus, as answered below, the actual number of possible placings of two persons in the queue indeed is $\,2\times 4=8$, but the total number of placings is 30, thus giving us a probability of $\,\displaystyle{\frac{8}{30}=\frac{4}{15}}\,$ .

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Yeah, right!!!! – Ashu May 26 '12 at 16:29

The answer is option 3.

How many options are there for the position of persons $A$ and $B$? In total there are $\binom 62=15$ possibilities. In how many of those options are these persons seperated by exactly one other person? In precisely 4 of them, namely when the are in position 1 and 3, or 2 and 4, or 3 and 5, or 4 and 6.

Edit: Maybe I should clarify my solution as there is apparently some confusion whether we took the interchangibility of $A$ and $B$ into account properly or not. Let me just consider the simpler version where we have only three people $A,B$ and $C$. Which possibilities are there?

$ABC$,$ACB$,$BAC$,$BCA$,$CAB$ and $CBA$. This are 6 options. The positions we are interested in are $ACB$ and $BCA$, so 2 of them. The probability is then $\frac 26$.

Another way to consider the same question is to ask ourselves which positions $A$ and $B$ may attend together. Then we have $(**-)$, $(*-*)$ and $(-**)$, where the stars mark $A$ and $B$ and the blank marks $C$. Out of these three options only $(*-*)$ is of the desired form, so the probability is $\frac 13$.

Similarly we can tackle the situation with a total of 6 people. We can either compute $\frac{8}{6\cdot 5}$, or $\frac{4}{\binom 62}$. Of course the result is the same, namely $\frac{4}{15}$.

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Whoops!! it was so simple... why didnt it strike me .. UGH!! ... thanks anyways. – Ashu May 26 '12 at 16:16
I think this answer doesn't take into account the two placements for A,B in each correct position (either A _ B or B _ A) , so the answer is, imo, (1)) – DonAntonio May 26 '12 at 16:20
@DonAntonio You have to divide by 2 at one point. I did it when I computed $\binom 62$, so I already took the interchangebility of $A$ and $B$ into account. Alternatively one could first compute $\frac {8}{30}$ and then divide by two. As a reality check: don't you think a probability of over $\frac 12$ seems to be too high? – Simon Markett May 26 '12 at 17:11
I think you've got a point, and address this in my added comment to my own answer. – DonAntonio May 26 '12 at 17:28
And your comment made me clarify my solution. Anyway this confusion helped improving both our solutions. – Simon Markett May 26 '12 at 17:37

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