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HSC Mathematics Marathon (6 Viewers)

Drongoski

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1-z=-2i sin(@/2)cis(@/2)
1-z^n= -2i sin(n@/2)cis(n@/2)

(1-z^n)/(1-z)=result
Shouldn't your 1st 2 lines be as modified above ??

You'll still get the same end result.

Whatever the answer to my query, still an elegant solution.
 
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deterministic

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New question:

How many ways are there to arrange 10 boys and 5 girls in a circle such that none of the girls are seated next to each other?

EDIT: it means that between any 2 girls, there must be at least 1 boy
 
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cutemouse

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New question:

How many ways are there to arrange 10 boys and 5 girls in a circle such that none of the girls are seated next to each other?
Consider the girls seated next to each other. This is done in 5! ways. This leaves 15-5+1=11 people units.

So No of ways girls sitting next to each other = 1*10!*5!

So No of ways girls not sitting next to each other = 1*14!-1*10!*5!=8.67*10^10 (3 sf).
 

cutemouse

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New question:

[maths]x^2-(p+iq)x+3i=0[/maths] (p, q are real) has roots [maths]\alpha[/maths] and [maths]\beta[/maths]. The sum of the square of the roots is 8.

(i) Write down expressions for the sum of roots and product of roots.

(ii) Hence find the possible values of p and q.
 

jyu

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Shouldn't your 1st 2 lines be as modified above ??

You'll still get the same end result.

Whatever the answer to my query, still an elegant solution.
Thanks
I checked my working, should be just 2i.
 

Pyrobooby

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New question:

[maths]x^2-(p+iq)x+3i=0[/maths] (p, q are real) has roots [maths]\alpha[/maths] and [maths]\beta[/maths]. The sum of the square of the roots is 8.

(i) Write down expressions for the sum of roots and product of roots.

(ii) Hence find the possible values of p and q.
(i) [maths]\alpha[/maths] + [maths]\beta[/maths] = p+iq,
[maths]\alpha[/maths][maths]\beta[/maths]=3i

(ii) p=3, q=1 or p=-3, q=-1.

Am I supposed to put the working out too?

Am I even correct :\
 

deterministic

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Consider the girls seated next to each other. This is done in 5! ways. This leaves 15-5+1=11 people units.

So No of ways girls sitting next to each other = 1*10!*5!

So No of ways girls not sitting next to each other = 1*14!-1*10!*5!=8.67*10^10 (3 sf).
Incorrect, perhaps I should clarify. No girls seated next to each other means a formation (assume a circle) BGGBGBGBBGBBBBB is invalid as 2 girls are next to each other. ie. between any 2 girls, there must be at least 1 boy.
 
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cutemouse

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Incorrect, perhaps I should clarify. No girls seated next to each other means a formation (assume a circle) BGGBGBGBBGBBBBB is invalid as 2 girls are next to each other. ie. between any 2 girls, there must be at least 1 boy.
In that case, it would be 1*6!*2!*7! = 7257600 I think...
 
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cutemouse

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I had another answer up there but I thought it was wrong (in the order of 10^10).

EDIT: Yeah I think my other answer was correct: 1*14! - (10!*5!+11!*4!+12!*3!+13!*2!) = 7.05*10^10 (3 sf) ... maybe not though.
 
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cutemouse

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Could you explain how you got these answers? Your answer is still not right.
If all 5 girls are together then no of ways that could happen is 1*10!*5!
4 girls ... 1*11!*4!
3 girls ... 1*12*3!
2 girls ... 1*13!*2!

total number of ways = 1*14!

So that's how I got that... But then again I only do General maths so what would I know...

EDIT: The above may be wrong since the 3, 4, 5 girls case contains the 2 girls case...
 
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jyu

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New question:

How many ways are there to arrange 10 boys and 5 girls in a circle such that none of the girls are seated next to each other?

EDIT: it means that between any 2 girls, there must be at least 1 boy
14!
 

cutemouse

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I don't think it's 14. I know at least 2 specific cases [BGBBGBBGBBGBBGB and BBGBGBGBGBGBGBG] with 48 and 6!*7! number of ways each...

EDIT: Oh 14! ... that's the total number of ways without restrictions.
 

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