What is the coefficient of the term $x^3y^5$, as a result of the binomial expansion of the following term? We have the term
$(1+xy+y^2)^n$ 
If we expand it using the binomial theorem, why is the factor of the term $x^3y^5$ the following: $4{n\choose 4}$? (The binomial coefficient n choose 4 multiplied by 4)

Yes, for n = 5, use the binomial theorem by substituting $xy + y^2$ for "$a$" and then expand $(1 + a)^5 = 1 + 5a + 10a^2 + 10a^3 + 5a^4 + a^5$.
Then substitute back $xy + y^2$ for $a$.........
Here, the only term which will give $x^3y^5$ is $5a^4$ whereby again using the binomial theorem on $5(xy + y^2)^4$ we get......
$5x^4y^4 + 5(4)x^3y^5 + 5(6)x^2y^6 + 5(4)xy^7 + 5y^8$
We get $20x^3y^5$ which is in keeping with $4\binom{5}{4}$
So to answer you question in a general sense, expanding a trinomial involves more than one application of the binomial theorem for some of the terms. 
