ruplun wrote:If '12' persons are seated at a round table, what is the probability that two particulars persons sit together?

12 persons can arrange themselves around the table in (12-1)!=11!;

as two particular persons sit together consider them as a one group, therefore we have 11 people to arrange around the round table which can be done in (11-1)!=10! also two particular person can exchange positions among themselves in 2! ways, hence required probability is 10!*2!/11!=2/11;

if instead of 12 there are 3 persons ? how to solve the prob

In this problem it is not even necessary to apply a formula.

Take any one person (person A) out of the twelve people and ask yourself how many possibilites there are that that person sits next to a particular person (person B). Person A sits next to Person B only when B sits directly to the right or directly to the left of A. There are 11 chairs on which B could sit, but only 2 chairs fulfil the condition of those two people sitting next to each other.

Hence, 2/11.

what will be the solution for the problem,
If among 3 persons 2 persons has to be seated next?
Ans : 2/2 = 1...please confirm

Well, that is true as long as those 3 people also sit at something round. Consider yourself sitting at a round table with your mom and dad. You will sit next to both of them, one will be to your right and one to your left. For them, the same holds true. So everyone will sit next to everyone and the probability of the situation you described is 100%.

if among 5 children there are 2 siblings , in how many ways can the children be seated in a row so that the 2 siblings do not sit together?

Please solve with detail explanation....