# Thread: Discrete and Combinatorial Math - Combination with Repetitions

1. ## Discrete and Combinatorial Math - Combination with Repetitions

I need a bit of help with a problem. I am pretty sure I know what formula I should use, I just don't know how.
Question:
Find the coefficient of $v^2w^4xz$ in the expansion of $(3v + 2w + x + y + z)^8$
I think I need to use $(a_1 + a_2 + ... + a_n)^r = \Sigma \binom{r}{x_1 x_2 ... x_n} a^{x_1}_1 a^{x_2}_2 ... a^{x_n}_n$ but I am not sure. Any help would be great! I have already looked thought the book and there isn't any other problems of this sort that I can find, and this book does not have a students solution manual.

2. In $(x_1+...+x_k)^n$ the coefficient of $x_1^{a_1}...x_k^{a_k}$ is $\frac{n!}{(a_1)!...(a_k)!}$.

This means the coefficent here is: $\frac{8!}{2!4!1!1!}$

EDIT: Wrong answer. Also, it should be $k=5$ and not $k=4$.

3. $\frac{8!}{2!4!1!1!}$ gives me 840. I have derive and when I expand $v^2w^4xz$ I get 120960 for the coeffiction. Are you sure about that solution?

4. Originally Posted by ThePerfectHacker
In $(x_1+...+x_k)^n$ the coefficient of $x_1^{a_1}...x_k^{a_k}$ is $\frac{n!}{(a_1)!...(a_k)!}$.

This means the coefficent here is: $\frac{8!}{2!4!1!1!}$
TPH has made an error here (and look, out the window, a flying pig).

The formula (called the multinomial theorem) is correct, but its application to your specific question needs tweaking: you have $3x_1$ rather than $x_1$ and $2x_2$ rather than $x_2$ .....

Think about the small changes that will make to the answer given by TPH.

5. I got it!
$840 \times 3^2 \times 2^4 = 120960$

Thank you guys so much, I finally understand this. Now I just wish you guys were here to help me go over my notes and help me study Oh well. Thank you. I will be coming back when I need help with things, this is a lot better help than I get from other people.