Binomial expansion question?

Write out the factorials in full e.g.

(n-7)! = (n-7)(n-6)(n-5)...(2)(1)

And then you should be able to do some cancelling.

Thanks for the tip,I tried it but I did something else wrong and didn't get the right answer,where am I going wrong? (Sorry if handwriting is a bit messy )

I don't have time right now to find your mistake but I recommend trying it again but keeping things factorised so don't work out 8!, 7!, 2^7 and 2^8 but consider cancelling instead. Your working will be simpler and lead to fewer mistakes. If you're still stuck let me know and I or someone else can check later.

I did it again and still got the same wrong answer. So there is something wrong with my method :/. Thanks for your help anyway.

$\dfrac{(n-8)!}{(n-7)!}$

This is not equal to $(n-8)$. Notice that $(n-7)$ is bigger than $(n-8)$.

If you expand (n-8)! Would it not be (n-8)(n-7)(n-6)....(n-1)(n)

So if it's divided by (n-7)! only (n-8) is left

If that's not the case then what should you get if you expand (n-8)! and (n-7)! (Sorry for being a nuisance )

If you consider e.g. 5! that is

5 x 4 x 3 x 2 x 1

The numbers get smaller by 1 each time

So (n-8)! is actually

(n-8)(n-9)(n-10)... x 3 x 2 x 1

Does that make sense?

That makes complete sense Thank you for explaining.I'll try the question again,I'll let you know if I get it right!

Sorry to bother again. In the markscheme it says that it should be (n-7)*2=8 But I seem to have it the other way around as 8*(n-7)=2