Binomial expansion question? Watch

FutureMissMRCS
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For question 6ii,I used the formula but don't know how to go about simplifying it to find the value of n?Can someone help please?
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FutureMissMRCS
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Notnek
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(Original post by FutureMissMRCS)
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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.
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FutureMissMRCS
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(Original post by Notnek)
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.
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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 )
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Notnek
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(Original post by FutureMissMRCS)
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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.
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FutureMissMRCS
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I did it again and still got the same wrong answer. So there is something wrong with my method :/. Thanks for your help anyway.
(Original post by Notnek)
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.
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Notnek
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(Original post by FutureMissMRCS)
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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).
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FutureMissMRCS
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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 )
(Original post by Notnek)
\dfrac{(n-8)!}{(n-7)!}

This is not equal to (n-8). Notice that (n-7) is bigger than (n-8).
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Notnek
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(Original post by FutureMissMRCS)
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?
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FutureMissMRCS
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(Original post by Notnek)
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!
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Notnek
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(Original post by FutureMissMRCS)
That makes complete sense Thank you for explaining.I'll try the question again,I'll let you know if I get it right!
It's a common mistake that happens the first time students do these types of questions - you probably won't make it again
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FutureMissMRCS
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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
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Notnek
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(Original post by FutureMissMRCS)
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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
Now you're saying that

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

But that's also wrong.
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