help with ruse Q (1 Viewer)

[mreditor16]

okay, I am so confused

can someone spell it out?

 

[Kurosaki]

For part a. it is relatively straight-forward- you have 6 lines, and for an intersection point to occur, you must choose 2 of them. 6C2=15.

For part b, first let's consider how many intersection points there are on each line. Clearly there are 5 right?
Let's call one specific line A. There are 5 other lines, and since no 3 lines or more are concurrent, there are 5 intersection points per line.

So then: choose a line. 6 lines, so we can do this in 6 ways. From the intersection points on that line, we select 3.
Sample space is 15C3, because we simply select 3 out of the 15 possible intersection points.


Is that helpful?

 

[mreditor16]

For part a. it is relatively straight-forward- you have 6 lines, and for an intersection point to occur, you must choose 2 of them. 6C2=15.

For part b, first let's consider how many intersection points there are on each line. Clearly there are 5 right?
Let's call one specific line A. There are 5 other lines, and since no 3 lines or more are concurrent, there are 5 intersection points per line.

So then: choose a line. 6 lines, so we can do this in 6 ways. From the intersection points on that line, we select 3.
Sample space is 15C3, because we simply select 3 out of the 15 possible intersection points.


Is that helpful?

yeh that makes sense. haha just so tired and I hate these point of intersection perms and combs Qs.

while you're at it, could you explain the last part:



thanks so much man. repped

 

[Kurosaki]

yeh that makes sense. haha just so tired and I hate these point of intersection perms and combs Qs.

while you're at it, could you explain the last part:



thanks so much man. repped

Call the lines A,B,C,D,E.
Let's pick one intersection point. There are 15 of them, so we can do this in 15 ways. Now, I'll arbitrarily let the lines that intersect there be called line A and B. Call the intersection point X.
This means that the next point you choose cannot be any point lying on line A OR line B. There are 5 points on line A including X, and 5 points on line B, also including X. So the number of points on lines A and B equals 5+5-1=9 (subtract 1 because X lies on both lines so you double counted that).

So now you have 15-9=6 points left to choose for the second intersection point, and then apply the same logic to find the number of points you can choose from for the last point.

You also need to divide by 3! because choosing, points X, Y and Z say is the same as choosing points Z, X and Y.

 

[Kurosaki]

It helps for these kinds of questions to draw a diagram .

 

[mreditor16]

It helps for these kinds of questions to draw a diagram .

thanks so much man for your help.

i'll read it through and then take another go.

 

[mreditor16]

wow that was tough to visualise.

but I did draw a diagram and that really helped.

thanks so much kurosaki got it out.

 

[Kurosaki]

That's good . What year is this paper? Seems hectic if it has combinatorics in the last question .

 

[mreditor16]

That's good . What year is this paper? Seems hectic if it has combinatorics in the last question .

http://4unitmaths.com/jamesruse2009.pdf

enjoy

 

[Kurosaki]

Legend, thanks mreditor16.

 

[Machiavelli1]

Ahah, found this Q in CSSA 1991

 

[IR]

Thought perms and combs was 3u. Hope Perms and Combs are swapped with something like harder mechanics.

 

[Carrotsticks]

Thought perms and combs was 3u. Hope Perms and Combs are swapped with something like harder mechanics.

It is, but this problem is best put under "Harder 3U" in Extension 2.

Also, I doubt Mechanics can get much harder to be honest. There's only so much you can do with high school methods without stepping out of the syllabus.

Though tbh I would not mind a conical pendulums combined with resisted motion problem (rotating within a medium of water anybody?).

 

[IR]

It is, but this problem is best put under "Harder 3U" in Extension 2.

Also, I doubt Mechanics can get much harder to be honest. There's only so much you can do with high school methods without stepping out of the syllabus.

Though tbh I would not mind a conical pendulums combined with resisted motion problem (rotating within a medium of water anybody?).

Would love to try that. Although not sure if in syllabus. More than 3/4 of the state however are not proficient with circular motion with variable angular velocity. However it hasn't been asked for a long time now. I doubt it will pop this year.

 

[axwe7]

Are there any answers to this paper?

 

[dan964]

Are there any answers to this paper?

yes I have them. check https://9eeba4054ee764a743f35bb25b5...papers_extension2.html?search=James Ruse 2009 for just 2009 James Ruse

https://9eeba4054ee764a743f35bb25b5.../Maths/trialpapers_extension2.html#James Ruse for all James Ruse X2 Trials