How many disciples does Tiffany have? (1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her disciples. (2) there are 28 ways that she could decide which 2 disciples she will recommend promoting.

Clearly the first statement alone is not sufficient. From the second statement: If there are n disciples it is given that nC2 = 28 i.e n(n-1)=56. So n = 8. Hence second statement alone is sufficient

sureshbala wrote:Clearly the first statement alone is not sufficient. From the second statement: If there are n disciples it is given that nC2 = 28 i.e n(n-1)=56. So n = 8. Hence second statement alone is sufficient What if OA is D, sir! Will you support that?

sanju09 wrote:sureshbala wrote:Clearly the first statement alone is not sufficient. From the second statement: If there are n disciples it is given that nC2 = 28 i.e n(n-1)=56. So n = 8. Hence second statement alone is sufficient What if OA is D, sir! Will you support that? Yes...the answer will be D. I misunderstood the first statement...Should get out of my office quickly.... Anyway.... From first statement if n is the number of disciples, the number of ways in which she can choose at least 2 disciples is 2^n - (nC0 + nC1) = 2^n - (n+1) Given 200<2^n -(n+1)<500 There is only one natural number satisfying this which is 8. Hence D will be the answer

By the way Sanju, nice to see you here as well...Hope you are doing well

sureshbala wrote:By the way Sanju, nice to see you here as well...Hope you are doing well I am sure you listen to Hindi songs... Aji rooth kar ab kaha.n jaaiyega Jaha.n jaaiyega hamei.n paaiyega Do you know Suresh, yours have been the quickest way to solve this problem, so far, though after realizing that you are in office, I am also leaving now, bye and good night!