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Help me with this probablity problem


Rakibul Raju

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  • 2 months later...

12 persons can sit along around a table in (12-1)! ways i.e 11!

Now considering that the 2 members who has to sit together as one unit, we have 11 units altogether, who can sit along in (11-1)! i.e 10! ways

 

Also the 2 persons sitting together can exchange their positions in 2! ways

 

So the probability is (10! * 2)/11! = 2/11

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  • 2 weeks later...
If ‘12’ persons are seated at a round table, what is the probability that two particulars persons

sit together?

We don't need to apply any counting techniques if we look at it this way: What is the probability that Ann and Bea sit together.

Let's just focus on Ann for a moment.

What is the probability that Bea is sitting beside her?

Well, there are 11 remaining seats and 2 of them are beside Ann.

So, P( Ann and Bea sit together) = 2/11

 

Cheers,

Brent

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  • 1 year later...
We don't need to apply any counting techniques if we look at it this way: What is the probability that Ann and Bea sit together.

Let's just focus on Ann for a moment.

What is the probability that Bea is sitting beside her?

Well, there are 11 remaining seats and 2 of them are beside Ann.

So, P( Ann and Bea sit together) = 2/11

 

Cheers,

Brent

Why is it that the number of ways 12 people can be seated at a round table is 11!? Why not 12!? explain please? and also the rest of the proces please.

 

Thanks

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Why is it that the number of ways 12 people can be seated at a round table is 11!? Why not 12!? explain please? and also the rest of the proces please.

 

Thanks

 

You will notice that I never said that 12 people can be seated at a round table in 11! ways (Chromepal made this claim). I'm not a big fan of "people sitting at a circle" questions, because the assumption is that the seats are identical. Since this is not stated in the question, I solved it in a way that does not require us to make this assumption.

 

Cheers,

Brent

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