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Find the coefficient of $x^{16}$ in $(x^{2} + x^{3} + x^{4} + x^{5})^{5}$

Please solve this question using Generating functions step by step ?

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x^4 x^5 ? addition in between ? why nt use latex ? easy to read
I think 135

Assuming polynomial as (x2+x3+x4+x5)5

we can rewrite it as (x2(1+x+x2+x3))5 = x10(1+x+x2+x3)5

now we have to find coefficient of x6 from (1+x+x2+x3)5 --------------( becoz x6 * x10 = x16 )

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

x1 + x2 + x3 + x4 + x5 = 6  such that 0 <= xi <= 3 where i can be 1 to 5

so complement of above equation will be atleast one of the xi 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+5-1C6 = 10C6 = 210

since least value greater than 3 is 4. That means the complement equation will have only one xi with least value 4

So consider x1 is greater than 3, that means 4 <= x1

rewriting the equation

x1 + 4 + x2 + x3 + x4 + x5 = 6

x1 + x2 + x3 + x4 + x5 = 2

number of ways of doing this = 2+5-1C2 = 6C2 = 15

similarly x2 or x3 or x4 or x5 can also be greater than 3

so total ways = 15 * 5 = 75

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

210 -75 =135 correct

we can use Pascal's triangle for finding this binomial expansion then higher order terms are neglected.

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this is what i was looking for as an explanation! thank you!