## Online Course Discussion Forum

### Help on II-A 6.20, 6.28, and 6.29

Help on II-A 6.20, 6.28, and 6.29

I don't understand how to use the binomial theorem, or things of that sort, on expressions that have more than 2 terms inside of them, so I don't really know how to do the questions mentioned in the subject without just grinding through the work for all of it.

Re: Help on II-A 6.20, 6.28, and 6.29

Imagine the Multinomial Theorem like this: You are arranging $x$'s, $y$'s, and $z$'s.

For example, in $(x+y+z)^3$, it's pretty obvious that every "term" (I use the word term very loosely here) can only contribute an $x$, a $y$, or a $z$ term.

Example 1: Find the coefficient of the $x^2 y$ term in $(x+y+z)^3$.

We can start by seeing that to get the $x^2 y$ term, we can get it by "choosing" 2 of the $(x+y+z)$ terms to "contribute" an $x$ to the product, and the other one "contributes" a $y$ term, so we get $\binom{3}{2}=3$ ways to get an $x^2 y$ term.

Example 2: Find the coefficient of the $xyz$ term in $(x+y+z)^3$.

We can start by seeing that to get the $xyz$ term, we can get it from "choosing" 1 of the terms to contribute an $x$, choosing 1 to contribute a $y$, and one to contribute a $z$.  Therefore, we can do this in $\binom{3}{1}\binom{2}{1}\binom{1}{1}=6$ ways, so the coefficient is $6$.

(You can also think of this as arranging $xyz$, with there being $\frac{3!}{1!1!1!}$)

Hope this helped!

Charles

Re: Help on II-A 6.20, 6.28, and 6.29

Thanks for this response, Charles!