actually,one approach we can apply here is combinatorics but it will work when n is very small.how to approach this problem
Pascal triangle
Here if you look closely every ith number is the binomial coefficient of the expansion (1+x)^k where k starts from 0.
Also number in line k at ith position = k!/( (k-i)! * i!)
and number in k line at (i-1)th position = k!/( (k-i + 1)! * (i-1)! )
Therefore number in line k at ith position is (number in k line at (i-1)th position) * (k - i + 1) / i
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