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MSI_10
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Please check attached files. I'm still trying to get used to how to explain maths properly. The transisiton to Uni is tough I have to admit.

Also for C would it be the same trick as B? So potentially the answer is 4 times 7 factorial? I'm not sure if my answer for B is right then that makes C too easy..

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brianeverit
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(Original post by MSI_10)
Please check attached files. I'm still trying to get used to how to explain maths properly. The transisiton to Uni is tough I have to admit.

Also for C would it be the same trick as B? So potentially the answer is 4 times 7 factorial? I'm not sure if my answer for B is right then that makes C too easy..

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If you think of placing Anita first and then you must consider two situations.
1. Anita placed at the end of the row
2. Anita placed anywhere except the end.
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MSI_10
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(Original post by brianeverit)
If you think of placing Anita first and then you must consider two situations.
1. Anita placed at the end of the row
2. Anita placed anywhere except the end.
I think I got it now..

for B is it 9! - 2(8!) and for C is it 9!-2(8!)-2(7!)? Not sure with C still...
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brianeverit
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(Original post by MSI_10)
I think I got it now..

for B is it 9! - 2(8!) and for C is it 9!-2(8!)-2(7!)? Not sure with C still...
the best method is probably to work out the number of arrangements with Anita and Charlene next to each other and subtract from your 9!
So how many ways can they be next to each other AC or CA, i.e. 2 ways
In how many different positions can the pair stand? (let this number be n)
How many ways can the remaining 7 be placed around them? (Let this number be m)
Then they can be stood next to each other in 2 x n x m arrangements.
Subtract from 9! and you have your answe4r.
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ClickItBack
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(Original post by MSI_10)
I think I got it now..

for B is it 9! - 2(8!) and for C is it 9!-2(8!)-2(7!)? Not sure with C still...
I agree with those answers, and think this is the simplest way to do the question.
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