Trinomial expansion coefficient

We have $(x + y + z)^6$ . If we think $(x + y) = a$ as a single term, we'll have $(a + z)^6$

In the final term we have $z^2$ , so it's reasonable that $\binom{6}{2}a^4z^2 = \binom{6}{2}(x + y)^4z^2$ will lead to the final term

But we don't need all the coefficients of $(x + y)^4$ , out of which we only need the coefficient of $x^2y^2$

The coefficient of $x^2y^2$ in $(x + y)^4$ is $\binom{4}{2}$

$\therefore$ The coefficient of $x^2y^2z^2$ in $(x + y + z)^6$ is $\binom{6}{2}\binom{4}{2}$ = 90

In order to achieve a product of $x^2y^2z^2,$ the " $x$ " must be "chosen" 2 times out of the 6. This gives $\binom{6}{2}$ combinations. Then, the " $y$ " must be "chosen" 2 times out of the remaining 4. This gives $\binom{4}{2}$ combinations. Then, the remaining factors will be " $z.$ " Thus, the coefficient of the $x^2y^2z^2$ term is:

$\binom{6}{2}\binom{4}{2}=\boxed{90}.$

For trinomial expansion, the general formulae can be used, that is - Coefficient of x^{a}.y^{b}.z^{c} in(x + y + z)^{n} can be given as (n!)/(a!).(b!).(c!)

Answering the above question within 5 seconds! (x + y + z)^{6}-- Coefficient of x^{2}.y^{2}.z^{2}={6!}{2!.2!.2!} =90