What is the co-efficient of $x^3y^4z^5$ in the expansion of $(xy+yz+zx)^6$? [duplicate]
I came across this problem and I was not able to think of a way to solve this. I tried to find a pattern by simplifying $(xy+yz+zx)^2$ and then extrapolating the results for the $6^{th}$ power but that also didn't help me. Can someone please tell me how to solve these kinds of problem?
Thanks in advance !!!
Solution 1:
write the expression in the form: $[y(x+z)+zx]^6$. The term with the 4th power of $y$ is: $$ \binom64 y^4(x+z)^4(zx)^2=\binom64 y^4(x^4+4x^3z +6x^2z^2+4xz^3+z^4)z^2x^2 $$ Thus the coeffecient of $x^3y^4z^5$ is: $$ 4\binom64 $$