Sum of coefficients in the expansion of $(2x+3y-2z)^n$ is $2187$ then the greatest coefficient in the expansion of $(1+x)^n$.
I put $x=y=z=1$ in the expansion. So, sum of coefficients became equal to $(2+3-2)^n=3^n$.
So, $3^n=2187\implies n=7$
Hence, greatest coefficient in expansion of $(1+x)^7$ is, ${}^7C_3={}^7C_4=35$
But I am not satisfied with this method. Is there another method to solve this without substituting some value of $x, y, z$?
The polynomial $(2x + 3y - 2z)^n$ will have a separate term for each solution to the following equation:
That is, by Stars and Bars theory, the polynomial will have $\displaystyle \binom{n+2}{2}~$ terms, which represents $a_1$ ranging from $0$ through $n$, and $a_2$ ranging from $0$ through $(n - a_1).$
For any corresponding solution represented by $~(a_1,a_2,a_3),~$ the corresponding term in the polynomial that represents
$$\text{coefficient} ~\times (2x)^{a_1} \times (3y)^{a_2} \times (-2z)^{a_3}$$
can actually be re-expressed as
$$\text{coefficient} ~\times \left[2^{a_1}3^{a_2}(-2)^{a^3}\right] \times x^{a_1} \times y^{a_2} \times z^{a_3} \tag1 $$
where
$$\text{(multinomial) coefficient} ~=~ \frac{n!}{a_1! \times a_2! \times a_3!}, \tag2 $$
The easiest way to compute the sum of the $~\displaystyle \binom{n+2}{2}~$ numbers that are collectively represented by (1) and (2) above is to set $(x,y,z) = (1,1,1),$ which forces each corresponding expression of $~\displaystyle x^{a_1} \times y^{a_2} \times z^{a_3}~$ to equal one.
Then, by the trinomial expansion of an expression of the form $(r + s + t)^n$, you have that the sum of the $~\displaystyle \binom{n+2}{2}~$ numbers that are collectively represented by (1) and (2) above is $(2 + 3 - 2)^n.$