What is the coefficient of the x⁴ y² z³ term of the trinomial expansion of: (2x-y+z)⁹

The best way to approach this question is to imagine (2x-y) as one variable. i.e. (2x-y+z)⁹ = ((2x-y) + z)⁹ , and then you can expand this expression binomially in the normal way to get something like this:
((2x-y) + z)⁹ = ₉ C₀ (2x-y)⁹ + ₉ C₁ (2x-y)⁸ z¹ + ... + ₉ C₃ (2x-y)⁶ z³ + ... + ₉ C₉ z⁹ .
So now all we need to find is the coefficient of x⁴ y² in (2x-y)⁶ , and by multiplying this by the coeff of z³ in the initial expression we will get the required number.

(2x-y)⁶ can be expanded in the usual binomial way:
(2x-y)⁶ = ₆ C₀ (2x)⁶ + ... + ₆ C₂ (2x)⁴ (-y)² + ... + ₆ C₆ (-y)⁶

so the coefficient of x⁴ y² z³ in (2x-y+z)⁹ = ₉ C_{3 6} C₂ 2⁴ (-1)² = (9!)/(3!6!) (6!)(2!4!) 2⁴

which simplifies down to 2⁶ x 3² x 5 x 7 = 20160.

There are algebraic expansion expanders on the internet which allow you to check this sort of thing, for example this one at quickmath.com

Hope this helps,

sam

Or you can use the fact that:

(a+b+c)n= å i+j+k=n  n!/(i!j!k!)ai bj ck

(This has a similar combinatorial proof to the usual binomial formula.) Then substitute a=2x, b=-y, c=z and the answer should come out immediately.