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Tried taking the interval to the first S as x1, from then to the next S as x2...etc. with sum being 9.

Possible partitions , including repeats, comes out as 211. This is the part where I am doubtful. Normally, I'd say ans would be (9!/4!) x 211, but the answer is given as (9!/4!) x 205.

what exactly is the error ? I am trying to assume x1 = 0, x2 = 1, x3 = 1 at which point 7 spaces remain , which can be partitioned in 7 ways among the remaining parts. An AP forms with 7,6,5.. for x3 > 1 which I solve for those cases using brute force to get 211....am I correct ?