Math Series #6

The given expression is, $(x+y-2z)^9=((x+y)-(2z))^9)$ Using binomial expansion we can write the above expression as, $((x + y) - (2z))^9 = \displaystyle \sum_{r=0}^9 {9 \choose r} (x+y)^{9-r}(-2z)^r$ Now we are looking for that expression containing the $z^4$ term. This can be obtained by keeping $r=4$ in the above expression, $2^4 {9 \choose 4} (x+y)^5z^4$ Again on applying binomial expansion and get, $2^4 {9 \choose 4} z^4 \displaystyle \sum_{r=0}^5 {5 \choose r} x^{9-r}y^r$ Now to get the term $x^3y^2$ we put $r=2$ , $2^4 {9 \choose 4}~{5 \choose 2} x^3y^2z^4$ On solving further we get the required term as $\boxed{20160x^3y^2z^4}$