Invitation to ALQUORA 2018

@manaranjanfav you have to just partition the entire array into 3 partition and then for that specific partition you’ll have the answer (1^x * 2^y * 3^z) where(x+y+z = n) and now the problem boils down to find a way to partition at and then add the corresponding number of ways to the answer. So, for now suppose we have only 2 values i.e. we need to partition the array into two half (suppose they are of the kind 2nd and 3rd). So, for a particular length n we have total of n+1 partitions possible (1st -> include 0 2nd type and n 3rd type, 2nd -> include 1 2nd type and n-1 3rd type and so on…). So, total number of ways for a particual length (such that we are allowed to use only 2nd and 3rd type) t is sum over all the above mentioned cases (2^0 * 3^t + 2^1 * 3^(t-1) + .....) and we know it’s a GP with sum (3^(t^+1^) - 2^(t^+1^)). Now, we know the number of ways we can make ornaments using 2nd and 3rd type and only case to include to get the answer is to include 1st type ornament and for that also we can do in same way. Suppose we have a total length of ‘n’ and so the total number of ways = sum of((0 length of 1st type and n length from remaining two) + (1 length of 1st type and (n-1) length of remaining two) + … ) = now as we know that (number of ways to select 1st type ornament of length x is 1) and so, we can reduce the above equation to the sum of expression (3^(t^+1^) - 2^(t^+1^)) from 0 to N.

I haven’t yet received any mail from u regarding bank details or anything,by when and how are u’ll gonna send the cash prize?And by when will i receive the laddus?

Hope this is not any kind of scam!!