Binomial Expansion is simple till this question...

What formula are you using for the binomial expansion? There are (at least) two versions commonly used, and for this problem it makes a difference.

If you know $\dfrac{n!}{(n-r)!r!}$ I am not sure why you cannot do the question

Do you know what $\dfrac{n!}{(n-r)!}$ cancels down to

if the question is (2+x/5)^10

we set it out like this:

1(2)^10
10(2)^9(x/5)
....
...

We haven't really learnt a formula just been taught how to do factorials on the calculator but I just taught myself this formula: nCr= n!/ (n-r)! r!

Do you know what n! means?

So what does the $x^2$ term of (1+x/2)^n look like when you do it this way?

So if n! = n(n-1)(n-2)(n-3)(n-4)........... etc

What would (n-2)! be and what would be left if you cancelled

this is what i've come up with:

1(1)^n=1n
n(1)^n-1(-3/2x)=n^n-1* -3/2x
?(1)^n-2(-3/2x)^2= ?

question mark because I tried using the formula to find nC2 and I got this n!/(n-2) 2! but I didn't know how to derive an answer

hmmm I think we'll have to write it out like this then:

n!/(n-2)! 2!= n(n-1)(n-2)...(n-n)/ (n-2)(n-1)* 2

Right or wrong ?

I am asking what you have left if you do $\dfrac{n!}{(n-2)!}$

or what do you get if you do $\dfrac{(n)(n-1)(n-2)(n-3).....}{(n-2)(n-3)......}$

I am asking what you have left if you do $\dfrac{n!}{(n-2)!}$

or what do you get if you do $\dfrac{(n)(n-1)(n-2)(n-3).....}{(n-2)(n-3)......}$

So far so good, and this is what TenOfThem has been addressing. You really do need to spot the pattern of the cancelling. (n-2)! gives all the factors of n! except for two factors - (what two factors?) - and these are all that is left in the numerator after(n-2)! is cancelled from top and bottom.