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Trinomial expansion
By Jason Wallace on Monday, March 25, 2002 - 03:10 am:

What is the coefficient of the x4y2z3 term of the trinomial expansion of: (2x-y+z)9

By Sam Davies on Monday, March 25, 2002 - 12:33 pm:


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 = 9C0(2x-y)9 + 9C1(2x-y)8z1 + ... + 9C3 (2x-y)6z3 + ... + 9C9 z9.
So now all we need to find is the coefficient of x4y2 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 = 6C0 (2x)6 + ... + 6C2 (2x)4(-y)2 + ... + 6C6 (-y)6

Þ the coefficient of x4y2z3 in (2x-y+z)9 = 9C3 6C2 24(-1)2 = (9!)/(3!6!) (6!)\(2!4!) 24

which simplifies down to 26•32•5•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

By Michael Doré on Monday, March 25, 2002 - 12:39 pm:

Or you can use the fact that:

(a + b + c)n = S 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.