Permtations and Combinations simplify the 4th term

Question #1 : Simplify the fourth term in
\(\displaystyle (x - 2y)^6\)

My answer: I used this formula..(or attempted I should say)

\(\displaystyle \L\\(a + b)^n :\:t_r = (\frac{n}{{r - 1}})a^{n - (r - 1)} b^( ^{r - 1)}\)


\(\displaystyle \L\\t_4 = (\frac{6}{5})x^3 (2y)^3\)


Question #2 which I have been working on but have got bogged down big time.

Find the numerical coefficient of the term containing \(\displaystyle x^4\) in :

\(\displaystyle \L\\\left( {x^2 + \frac{2}{X}} \right)^8\)

My answer or start

\(\displaystyle \L\\\begin{array}{l}
(x^2 )^8 ,(x^2 )^7 ,(x^2 )^6 \\
x^{16} ,/:x?,/:x? \\
\end{array}\)

I realize once you can figure out this Binomial Theorem Pattern then you can just use the formula in question Number one. My books are absolute crap and I really need some easier, simpler explanation.

Thank you for helping
Louise

What I am about say really does not help you.
I have written option pieces against this particular question for several years.
The fact is there may be several correct answers: it depends upon the definition of order. That is subjective. Actually ‘the fourth term’ can mean any among seven! I can prove that.

Your question #2 is well defined!
Any term in that expansion looks like
\(\displaystyle \L\left( {x^2 } \right)^J \left( {2x^{ - 1} } \right)^k = \left( 2 \right)^k Cx^{2J - k} \quad k + j = 8,\quad \& \quad C = \frac{{8!}}{{\left( {k!} \right)\left( {j!} \right)}}\)

Wow I am gonna need some serious help. If anyone is reading this and you know of any and all online resources that may be able to explain Binomial Theorem please feel free to post them!!
Thanks
Louise

UPDATE:
I have been doing some online research and I think the answer for number one might look more like:

\(\displaystyle \L\\(x - 2y)^6 = {}_6C_0 (x)^6 ( - 2y)^0 + {}_6C_1 (x)^5 ( - 2y)^1 + {}_6C_2 (x)^4 ( - 2y)^2 + {}_6C_3 (x)^3 ( - 2y)^3\)

The fourth term being \(\displaystyle \L\\{}_6C_3 (x)^3 ( - 2y)^3\)

The simplified something like \(\displaystyle \L\\20(x)^3 - 8y\)

And for Question #2
wouldn't they be asking for the fifth term? Would it look something like \(\displaystyle \L\\\begin{array}{l}
8(x^2 )^4 \left( {\frac{2}{x}} \right)^4 \\
4 \\
\end{array}\)
Then the numerical coefficient of that term would be \(\displaystyle {}_8C_4 = 70\)

I really hope all this work doesn't go to waste and somebody can make some sense out of it.
Thank you
sincerly
Louise

The Binomial theorem is pretty trivial, perhaps it is not explained well in school. The coefficient of \(\displaystyle a^{k}b^{n-k}\) in the expansion of \(\displaystyle (a+b)^{n}\) is the number of ways you can pick k a's and n-k b's from the n factors
(a + b). This is, of course, \(\displaystyle {n \choose k}\)