find the Coefficient of term x^{2}y^{3} in below expansion:
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...}
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$\displaystyle x^2y^3$