Assuming polynomial as (x^{2}+x^{3}+x^{4}+x^{5})^{5}

we can rewrite it as (x^{2}(1+x+x^{2}+x^{3}))^{5} = x^{10}(1+x+x^{2}+x^{3})^{5}

now we have to find coefficient of x^{6} from (1+x+x^{2}+x^{3})^{5} --------------( becoz x^{6} * x^{10} = x^{16} )

now we can imagine above polynomial as 5 boxes ( becoz whole power is 5 ) and each box can have either 0 or 1 or 2 or 3 items in it and final addition of all items of the boxes will be 6. That means

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 6 such that 0 <= x_{i} <= 3 where i can be 1 to 5

so complement of above equation will be atleast one of the x_{i} will have value greater than 3

so to find this first we will have to find total ways of doing this and then deduct those ways which include items greater than 3 in the boxes.

total ways = ^{6+5-1}C_{6} = ^{10}C_{6} = 210

since least value greater than 3 is 4. That means the complement equation will have only one x_{i} with least value 4

So consider x_{1} is greater than 3, that means 4 <= x_{1}

rewriting the equation

x_{1} + 4 + x_{2} + x_{3} + x_{4} + x_{5} = 6

x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 2

number of ways of doing this = ^{2+5-1}C_{2} = ^{6}C_{2} = 15

similarly x_{2} or x_{3} or x_{4} or x_{5} can also be greater than 3

so total ways = 15 * 5 = 75

So required answer = total ways - complemented ways = 210 -75 =135