If n person are seated at a round table then the probability that m particular person sit together is howmuch

take m as a 1 person,
so permutation=(n-m+1)!
m can be arranged as m!
total arrangement=m!*(n-m+1)!
sample space=(n-1)!
ans: ((n-m+1)!*m!)/(n-1)!
11 years agoHelpfull: Yes(21) No(3)
I think (n-m)!*m!
11 years agoHelpfull: Yes(2) No(3)
@bhumi...its a round table so that u fixed a 1st sit ie 1 person,therefore
(n-m+1-1)!*m!/(n-1)! is the r8 ans......ANANYA u r r8!!!!
11 years agoHelpfull: Yes(2) No(0)
circular permutation=n-1!
first we will fix m persons as 1.
now we are left with n-m persons.
don't forget to add m persons as 1 in n-m.
so we have (n-m+1-1)!/(n-1)!
11 years agoHelpfull: Yes(2) No(1)
Ananya you are right
11 years agoHelpfull: Yes(1) No(0)
i dont noe if i'm right but for circular arrangement if no of permutations is (n-1)! then for this particular case we we'll have (n-m-1)! in which m elements can also further be arranged in m! ways so the answere should ne (n-m-1)! m!

@ananya and bhumi why have you devided the answer with (n-1)!
11 years agoHelpfull: Yes(1) No(0)
total number of space available for (n-m) artical= n-m+1;
we have choice to put one group(containing m member) at (n-m+1)space & m member can also be rearranged in themselfe so;
so tatal way=(n-m+1)P1*m!={(n-m+1)!/(n-m+1-1)!}*m! = {(n-m+1)!/(n-m)!}*m!
11 years agoHelpfull: Yes(1) No(1)
(((n-m)+1)!*m!)/n!
11 years agoHelpfull: Yes(0) No(3)
@aishwarya
because the qn is abt probability:favourable events /all possible events
11 years agoHelpfull: Yes(0) No(0)
(n-m)!*m!/(n-1)!
11 years agoHelpfull: Yes(0) No(0)

If n person are seated at a round table then the probability that m particular person sit together is howmuch?