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Everything posted by m00li

  1. Let S1=(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11) S2=(n-1)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11) S3=(n-1)(n-2)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10)(n-11) so on till S11=(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)(n-9)(n-10) where n is any natural number then the sequence is given by (4/3628800)S1 -(8/362880)S2 +(16/80640)S3 -(26/30240)S4 +(41/17280)S5 -(57/14400)S6 +(79/17280)S7 -(104/30240)S8 +(138/80640)S9 -(184/362880)S10 +(241/3628800)S11
  2. you posted the answer too! realized now that ts an old problem
  3. Yes Bonanova, I did not permit zeros as I thought in that case Rainman would have given a different initial reply (that he can never find out Y-San's product)
  4. Fifty school children lined up at random in five rows and ten columns. No two of the children have the same birthday. What is the probability that the youngest child among the oldest children in each column is NOT older than the oldest child among the youngest children in each row ?
  5. In any event, I'm marking the puzzle solved. Good work. First of all, thanks for the star. Its been decades since I got one Now,..
  6. Since, Rainman knows the product, he doesn't even have to think about the other number. This doesn't change the fact that there are indeed two answers to your problem. Or, am I wrong in stating that there are two possible answers to your question?
  7. a 3 legged stool with vertically straight legs of heights 10 meter,1 meter and 1/2 meter will be very unstable. do u mean to say all three legs are equal? if so, then i do not understand the question
  8. I didn't do it, I just remembered it
  9. thats why i wrote that it needs some refining as theoretically speaking you cannot apply the same strategy any more but have to modify it now,
  10. How can you mirror player 1's first move in the center of paper? If player 1 places their point at (0, 0), place yours at (0, 1/infinity).
  11. question looks incorrect. the informants are providing info only about who was in which room, so we can at best deduce who was where. The question assumes that by knowing who was where, we will know the murder. The only relationship between a murderer and location information is that the murderer was alone in his room. This must imply that the murderer is the ONLY person who is alone in a room, otherwise we cannot deduce the murderer from the location information alone. Having said that, additional information on why I think so:
  12. More recursions: Let f(t,m) denote the number of ways of arranging m colored balls in t slots, where t >= m and supply of each m coloured ball is infinite. Then final answer is nCmf(t,m) / nt Here are the different recursions I have come up with for f(t,m) (alas, no direct formula in m,n) 1) mt - mC1f(t,1) - mC2f(t,2) - mC3f(t,3) .... - mCm-1f(t,m-1) 2) tC1f(t-1,m-1) + tC2f(t-2,m-1) + tC3f(t-3,m-1) + ... + tCt-m+1f(m-1,m-1) 3) mf(t-1,m-1) + m2f(t-2,m-1) + m3f(t-3,m-1) + ... + mt-m+1f(m-1,m-1) 4) m( f(t-1,m) + f(t-1,m-1) ) where f(t,1) = 1, f(t,2) = 2t-2, and f(t,t) = t!
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