Really shit at Prob/perms/combs (1 Viewer)

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I've been just shit at probability/perms/combs all my life (Q5 of 2011 2U paper baby).

Any tips/tricks/worksheets I can get my hands on? We're not doing it in class because my teacher is a **** but it features in every HSC paper...

Thanks
 

Beaconite

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lol same here! I suck at those topics.....But i did a lot of questions, im getting better! :D
 

Shadowdude

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see if you guys came to my seminar partially on enumeration and combinatorics at unsw open day...


Anyways, I still have the handout to that - so if you guys want to PM me your email addresses, I, or someone else in MATHSOC, can send you a copy of those notes over. They cover understanding the question, setting out the working logically, and smart ways to do the questions for Perms/Combs/Probability. (also have a few worked examples and extra questions for you to try)

They also contain stuff on partial fractions (Heaviside methods), integration by substitution, calculator tricks, general hints and tips, and L'Hopital's Rule for checking results of limits.


Sincerely,
Shadowdude

your friendly local unsw mathsoc representative
 
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Sindivyn

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If you can't solve a question and it has a relatively low sample space or repetitions, draw a tree diagram or table of values.
Consider working with the converse event if the event is complimentary - it's often a LOT easier this way.

Also, look at the probability question at the start of the paper and let it sit with you, it is likely to click while you're doing another question (same as circle geometry).
 

jeffwu95

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see if you guys came to my seminar partially on enumeration and combinatorics at unsw open day...


Anyways, I still have the handout to that - so if you guys want to PM me your email addresses, I, or someone else in MATHSOC, can send you a copy of those notes over. They cover understanding the question, setting out the working logically, and smart ways to do the questions for Perms/Combs/Probability. (also have a few worked examples and extra questions for you to try)

They also contain stuff on partial fractions (Heaviside methods), integration by substitution, calculator tricks, general hints and tips, and L'Hopital's Rule for checking results of limits.


Sincerely,
Shadowdude

your friendly local unsw mathsoc representative
Hey i came haha 30 minutes was not enough hahah!
 

Shadowdude

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Hey i came haha 30 minutes was not enough hahah!
I actually ran over time... you might've seen the other execs giving me hand signals to hurry up.


but it was good though, right? helpful, at least?

The take away from that session - from the enumeration part was: "Think clearly, work logically".
 

jenslekman

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see if you guys came to my seminar partially on enumeration and combinatorics at unsw open day...


Anyways, I still have the handout to that - so if you guys want to PM me your email addresses, I, or someone else in MATHSOC, can send you a copy of those notes over. They cover understanding the question, setting out the working logically, and smart ways to do the questions for Perms/Combs/Probability. (also have a few worked examples and extra questions for you to try)

They also contain stuff on partial fractions (Heaviside methods), integration by substitution, calculator tricks, general hints and tips, and L'Hopital's Rule for checking results of limits.


Sincerely,
Shadowdude

your friendly local unsw mathsoc representative
i like your lecture style - it was wonderful.
 
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I do have your notes! They are very helpful...it's probably just me - intrinsically **** at this topic ==
 

Shadowdude

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yea and the notes are good :L
Thanks! Glad you liked them, we... were all stressing on getting them done on time.

I do have your notes! They are very helpful...it's probably just me - intrinsically **** at this topic ==
Thanks asianese! However, I have to ask - weren't we all bad at the start?

I think the tools I've given you in the sheet would form a better foundation now. Like, I struggled at that in high school, like really struggled - and if you were at the talk, you would've heard me say that the methods I got taught in high school are... terrible.

But I learned a new and better method at uni, which lo and behold, anyone can use! It takes longer to write down, but it'll get you a correct answer more often because it'll limit the silly mistakes, and also highlight them easier when you make them.

Of course, there's no magic fix - there's a lot of hard work and effort and time that must be spent on learning the craft, and developing your skills. At these higher echelons of maths, you don't just "get it". You have to work a bit to understand the material.

But along the way - I think you'll find the notes will be a big help to you in conquering the questions!
 

Carrotsticks

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Asianese, I have an idea.

Make a thread purely for you and Perms/Combs questions, and answer a couple of questions explaining your logic.

This way, we can have some insight as to how your thinking processes work, and whether you are using the best method to do it or not.

If you get stuck or if we see that there exists a short (and more logical presumably) way, then we can tell you.
 
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That's a great idea. thanks!

I've got a list of all the HSC questions relating to this, so i'll start with those.
 

lolcakes52

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Hey shadowdude, I went to the seminar as well and it was great. One of the questions in the book, I think the one about how could a group of people be arranged in 3 rooms, might have been incorrect. In the coroneos past paper book the question is done differently and has another answer.... I think. But still, the seminar was amazing and very useful.
 
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Ok some attempts.

1983 7i)
A city council consists of 6 Liberal and 5 Labor aldermen, from whom a committee of 5 members is chosen at random. What is the probability that the Liberals have a majority on the committee?

_____

Pr(lib majority) = Pr ( 3+4+5 liberals)

1. ways of picking 3 libs and 2 lab ... C(6,3)C(5,2)
2. ways of picking 4 libs and 1 lab ... C(6,4)C(5,1)
3. ways of picking 5 libs and no labs ... C(6,5)

ways of picking any committee of 5 ... C(11,5)

therefore Pr(majority libs) =

1985 8i)
a) In how many ways can 4 persons be grouped into two pairs to play a set of doubles tennis?
b) The eight members of a tennis club meet to play two simultaneous sets of doubles tennis on two separate but otherwise identical courts. In how many different ways can the members of the club be selected for these two sets of tennis?

_______

I drew a court lol anyway

a)
1. Choose first pair...C(4,2)
2. Choose 2nd pair...C(2,2)
3. Each side is identical so divide by 2!

Ways =

b)
I drew 2 courts.
1. Choose 1st pair ... C(8,2)
2. Choose 2nd pair ... C(6,2)
3. Choose 3rd pair ... C(4,2)
4. Choose last pair ... C(2,2)
5. Each court is identical ... divide by 2!
6. Each side is identical ... divide by 2!

Ways =

1986 8ii)
A committee of 4 women and 3 men are to be seated at random around a circular table with 7 seats. What is the probability that all the women will be seated together?

____

Consider a table with all of them.
1. Fix one man.
2. Group women together
3. Arrange 2 un-fixed men and the group of women ... 3!
4. Order the women (lol)... 4!
5. Total ways of arranging women together = 4!3!
6. Total ways of arranging 7 people ... 6!
7. Pr(women together) =

Don't judge me :(
 
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barbernator

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Ok some attempts.

1983 7i)
A city council consists of 6 Liberal and 5 Labor aldermen, from whom a committee of 5 members is chosen at random. What is the probability that the Liberals have a majority on the committee?

_____

Pr(lib majority) = Pr ( 3+4+5 liberals)

1. ways of picking 3 libs and 2 lab ... C(6,3)C(5,2)
2. ways of picking 4 libs and 1 lab ... C(6,4)C(5,1)
3. ways of picking 5 libs and no labs ... C(6,5)

ways of picking any committee of 5 ... C(11,5)

therefore Pr(majority libs) =

1985 8i)
a) In how many ways can 4 persons be grouped into two pairs to play a set of doubles tennis?
b) The eight members of a tennis club meet to play two simultaneous sets of doubles tennis on two separate but otherwise identical courts. In how many different ways can the members of the club be selected for these two sets of tennis?

_______

I drew a court lol anyway

a)
1. Choose first pair...C(4,2)
2. Choose 2nd pair...C(2,2)
3. Each side is identical so divide by 2!

Ways =

b)
I drew 2 courts.
1. Choose 1st pair ... C(8,2)
2. Choose 2nd pair ... C(6,2)
3. Choose 3rd pair ... C(4,2)
4. Choose last pair ... C(2,2)
5. Each court is identical ... divide by 2!
6. Each side is identical ... divide by 2!

Ways =

1986 8ii)
A committee of 4 women and 3 men are to be seated at random around a circular table with 7 seats. What is the probability that all the women will be seated together?

____

Consider a table with all of them.
1. Fix one man.
2. Group women together
3. Arrange 2 un-fixed men and the group of women ... 3!
4. Order the women (lol)... 4!
5. Total ways of arranging women together = 4!3!
6. Total ways of arranging 7 people ... 6!
7. Pr(women together) =

Don't judge me :(
for the tennis ones, I think there is an easier and more efficient way IMO.

a) set 1 person to be paired. There is 3C1 ways of picking a partner, the other pair comes as a result, hence 3 ways
b) again, set 1 person, 7 partners, next set another, 5 partners, again another, 3 partners, and then youre set. 7x5x3=105. now the teams can play in 3 different arrangements, so the answer is 315. your answer is 630 i think you may have missed something not sure, or i could be wrong.

i will look at some others later.
 

deswa1

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1986 8ii)
A committee of 4 women and 3 men are to be seated at random around a circular table with 7 seats. What is the probability that all the women will be seated together?

____

Consider a table with all of them.
1. Fix one man.
2. Group women together
3. Arrange 2 un-fixed men and the group of women ... 3!
4. Order the women (lol)... 4!
5. Total ways of arranging women together = 4!3!
6. Total ways of arranging 7 people ... 6!
7. Pr(women together) =

Don't judge me :(
1. Group women together. Arrange everyone around the table (3 men and 1 superwoman) -> 3!
2. Arrange women- 4!
3. Total arrangements- 6!
4. Therefore probability= (4!3!/6!)

I don't think you need to fix the man etc
 

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