m00li

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

Nicely done

you posted the answer too! realized now that ts an old problem

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)

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 ?

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,..

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?

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

I didn't do it, I just remembered it