Re: Combinatorical question

On Feb 20, 2:53 pm, "Simon Johan" <si...@xxxxxxxxxxxxx> wrote:
"Mariano Suárez-Alvarez" <mariano.suarezalva...@xxxxxxxxx> wrote in message

news:f2f324c8-e5fe-4fdb-9cec-4b18e78cd966@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



On Feb 20, 8:06 pm, "Simon Johan" <si...@xxxxxxxxxxxxx> wrote:
Hi,

I want to calculate the probability that two people, say Bob and Jane, in
a
bus with 14 people will sit next to each other if the people are being
seated randomly.

My idea is the calculate how many ways Bob and Jane can sit next to each
other divided by the total number of ways people can be placed in the
bus.
The total number of ways 2 people can be placed next to each other among
14
people is 14! / (14 - 2)! = 182 ways. In how many of these do Bob and
Jane
sit next to eachother?

The solution says that the probability is 1/13, but why? This implies
that
Bob and Jane can sit next to each other in 14 of the possibilities, but
why?

The answer very much depends on how the seats are placed
in the bus, what `sit next to each other' means, and
how many seats there are...

There are 14 seats in the bus. I'm sorry for not mentioning that in my first
post. The problem does not say how the seats are placed, but I assume that
people sit together two and two.

Here is a slightly different version (not what you wanted, but maybe a
nice homework problem, depending on the course level): suppose the
people do not sit 2-by-2, but just individually in the 14 seats that
are arranged in a single row. So, if Bob sits in seat 1 there is just
one seat next to him, and similarly for seat 14. However, if he sits
in seats 2--13 there are two seats next to him. What is the answer to
your question in this case?

R.G. Vickson
.

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