find the Coefficient of termxin below expansion:^{2}y^{3}

expansion :(2+x-y+3z)^{10 }^{please write a good description for solution* or how to solve it? * for a person who is a dummy in math and combinatorics...}

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- Feb 28th 2013, 12:40 AMmshouCoefficient and expansions.
find the Coefficient of term

**x**in below expansion:^{2}y^{3}

expansion :**(2+x-y+3z)**^{10 }^{please write a good description for solution* or how to solve it? * for a person who is a dummy in math and combinatorics...} - Feb 28th 2013, 01:09 AMjakncokeRe: Coefficient and expansions.
80,640. Use the multinomial theorem which says $\displaystyle (x_1+...+x_r)^n = \sum_{P(n)}{n \choose k_1,k_2,..,k_r } * (x_1)^{k_1}...(x_n)^{k_r}$(Basically sum over all non negative integer partitions of n.)

the $\displaystyle {n \choose k_1,k_2,..,k_r} = \frac{n!}{k_1!...k_r!}$

basically, it says the powers on the exponents of each term you have in your parenthesis, namely 2,x,y,z must add up to 10.

so how you can write 10 = 0+0+3+7 which is equivalent to $\displaystyle \frac{10!}{0!0!3!7!}2^0*x^0*y^3*z^7$

so since you asked for $\displaystyle x^2y^3$ this means since we don't see z so the power of z is 0. 2+3+0+x = 10 and so x = 5 (which is the power 2 is raised to)

so you have $\displaystyle \frac{10!}{0!5!3!2!}2^5*x^2*y^3*z^0$ = 80640$\displaystyle x^2y^3$ - Feb 28th 2013, 01:25 AMMINOANMANRe: Coefficient and expansions.
It is - 80640 x^2y^3

MINOAS - Feb 28th 2013, 01:28 AMmshouRe: Coefficient and expansions.
well. thanks and a question:

isn't there any difference between for instance 3x and x, and -y and y ? - Feb 28th 2013, 01:43 AMmshouRe: Coefficient and expansions.
well. thanks and a question:

isn't there any difference between for instance 3x and x, and -y and y ?