Question

6 person are seated at a round table. What is the probability that 2 particular persons sit together?

Answer 1

call the two particular persons, A, B

Answer 2

@perl

Answer 3

@ganeshie8 . Number of ways 6 persons sit on a
round table = (6 - 1)! = 5!
.-. n(S) = 5!
Since two persons always sit together, we con-
sider them one. So, total persons are 4 + 1 = 5
.•. Number of ways these 5 persons sits on the
round table is (5 - 1)! = 4! ways and these two
persons sit in 2! ways among themselves. N(E)=4!*2!
N(E)/N(S)=2/5

Answer 4

put A and B in the box
`AB`

Answer 5

now total persons :-
`AB`P1,P2,P3,P4
total number of persons=5

Answer 6

arrange them in a circle :-
(n-1) !=((5-1)!

Answer 7

total outcomes :- (6-1) !

Answer 8

and we can arrange AB in two ways like AB and BA
so,multiply by 2 !
so,we get 4 !*2!

Answer 9

\[\large \bf probability=\frac{4! \times 2!}{5!}\]

Answer 10

why we taking AB in one box.that is confusing meand why we are taking circular seating for (n-1)!

Answer 11

i mean the two particular persons be A and B and they want to sit together so i put them in box or simply you can say that they never separate

Answer 12

REMEMBER :-
In circular permutation,we arrange `n` people in `(n-1) !`
and in linear or simple permutation,we arrange `n`people in `n!`

Answer 13

hope you understand @dg2