Perfecting Permutations and Combinations (1 Viewer)

[Sy123]

So I have come to the realisation that the only thing weighing me down at all is Permutations and Combinations.
And I want to perfect my ability to do the harder problems in Perms/Combs

I find that for the harder problems, I usually end up not seeing something, end up double counting or not taking into consideration very minor and little things which change the result drastically.

So how do I improve my approach to these problems? Practise doesn't work because I'll just keep doing the problems incorrectly.

Are there maybe any good online resources and stuff to try and approach them properly? i.e. khanacademy etc

Thanks.

 

[lochnessmonsta]

This is my weakness too. And I understand the 'practice doesnt really help that much' bit, because of the sheer number of possibilities of questions, and that each question looks essentially unique to me. I believe there's a weakness in my understanding of the theory, and not even my teacher can help me.

 

[asianese]

Lol Seems like yesterday: http://community.boredofstudies.org/showthread.php?t=290117

 

[asianese]

I refer you to this document: http://www.mediafire.com/?mlxfcg9imjscc4d

Compiled last year for practice.

 

[braintic]

I think you just have to keep asking questions here, and see if you can find more connections between apparently different problems.
For example, Ext 2 HSC 1992 Q7a. It is so much easier once you see it is isomorphic to the problem of forming words from the letters RRRRDDDDDD.

Here is a question:

How many solutions consisting of non-negative integers does the equation a+b+c+d=50 have?
(For example a=12, b=13, c=0, d=25 is one solution)

It is actually very easy, but only once you relate it to something more concrete. Apologies if you've already seen this one.

Please could others refrain from jumping in.

 

[Sy123]

I refer you to this document: http://www.mediafire.com/?mlxfcg9imjscc4d

Compiled last year for practice.

Thank you.

I think you just have to keep asking questions here, and see if you can find more connections between apparently different problems.
For example, Ext 2 HSC 1992 Q7a. It is so much easier once you see it is isomorphic to the problem of forming words from the letters RRRRDDDDDD.

Here is a question:

How many solutions consisting of non-negative integers does the equation a+b+c+d=50 have?
(For example a=12, b=13, c=0, d=25 is one solution)

It is actually very easy, but only once you relate it to something more concrete. Apologies if you've already seen this one.

Please could others refrain from jumping in.

I see what you mean by simplifying the problem down. For example, if we split 12 people into 3 groups of 4, the ways to do so is:

right? (I think)

But what if 2 people cannot be in the same group? Would it be, first putting those people in the group:



Then selecting people to put in after that:

is this correct?

===

As for your problem, is a good way to simplify it, to consider when none are zero. When a=0, a=b=0 and a=b=c=0?

I can do when a=b=c=0 (obviously) and a=b=0 (26) but I don't know how to a=0 elegantly.
I am thinking of considering the 48 cases for what b could be. And then getting a series to sum up....but I don't think that is what you're looking for.

EDIT: I think I'll watch all of khanacademy's probability playlist, starting from scratch....

 

[braintic]

As for your problem, is a good way to simplify it, to consider when none are zero. When a=0, a=b=0 and a=b=c=0?

I can do when a=b=c=0 (obviously) and a=b=0 (26) but I don't know how to a=0 elegantly.
I am thinking of considering the 48 cases for what b could be. And then getting a series to sum up....but I don't think that is what you're looking for.

EDIT: I think I'll watch all of khanacademy's probability playlist, starting from scratch....

Regarding my problem, you don't need to consider cases.
Imagine lining up 50 tokens, and then placing three dividers between them to divide them into 4 groups. Two or more dividers can be adjacent corresponding to zero solutions.
Can you see that all different arrangements of the 50 tokens and 3 dividers correspond to all different solutions to the given equation?
Answer ...?

(Regarding your other question, I'm just taking a break from doing other work, so I'll get back to it later.)

 

[Sy123]

Regarding my problem, you don't need to consider cases.
Imagine lining up 50 tokens, and then placing three dividers between them to divide them into 4 groups. Two or more dividers can be adjacent corresponding to zero solutions.
Can you see that all different arrangements of the 50 tokens and 3 dividers correspond to all different solutions to the given equation?
Answer ...?

(Regarding your other question, I'm just taking a break from doing other work, so I'll get back to it later.)

Ah yep I see that, clever, thanks.

right?

Because there are 53 elements to arrange, 3 dividers 50 tokens, and they are not distinct.

 

[asianese]

Check 2012 HSC Q16 ai, ii)

 

[braintic]

Ah yep I see that, clever, thanks.

right?

Because there are 53 elements to arrange, 3 dividers 50 tokens, and they are not distinct.

That's right. But I prefer to think of it as 53C3. That is, you are picking 3 places from 53 available to place the 3 dividers. Exactly the same thing though.

I think the message is, when you get stuck try to find an isomorphic problem that you can solve.

 

[Trebla]

A possibly useful approach for some problems is to look at the most specific example of the specified arrangement and 'multiply out' each dimension of specificity accordingly to get the total count

 

[hayabusaboston]

This is my weakness too. And I understand the 'practice doesnt really help that much' bit, because of the sheer number of possibilities of questions, and that each question looks essentially unique to me. I believe there's a weakness in my understanding of the theory, and not even my teacher can help me.

I c wat u did ther...

 

[asianese]

I c wat u did ther...

== no.

 

[VBN2470]

The whole topic itself can be very difficult as there is such a broad range of questions. However, getting the technique in answering questions is most important when arriving at a solution. I'd say the best way is to expose yourself to as many types of permutations/combinations questions and keep practicing. There is only a set difficulty they can go up to so I'd say practicing lots of questions especially the harder ones really helps. A good book for this is Terry Lee 3U (Fundamental) and the Harder 3U Topics Chapter in the 4U (Advanced) book as well as it covers some very useful techniques. Also past papers helps heaps. Sometimes people get lucky with these types of questions but this topic is never heavily tested in either 3U/4U, only probably a maximum of 2 - 3 marks they can ever ask in the HSC paper (from what I've seen) and sometimes they don't even ask. So don't worry too much about it, everyone is left with uncertainty when answering a question in an exam and perfecting this topic is not as easy as other topics. Practicing is probably the best way to increase your chance to get a question right as well as understanding and setting up a given situation.

 

[Makematics]

The whole topic itself can be very difficult as there is such a broad range of questions. However, getting the technique in answering questions is most important when arriving at a solution. I'd say the best way is to expose yourself to as many types of permutations/combinations questions and keep practicing. There is only a set difficulty they can go up to so I'd say practicing lots of questions especially the harder ones really helps. A good book for this is Terry Lee 3U (Fundamental) and the Harder 3U Topics Chapter in the 4U (Advanced) book as well as it covers some very useful techniques. Also past papers helps heaps. Sometimes people get lucky with these types of questions but this topic is never heavily tested in either 3U/4U, only probably a maximum of 2 - 3 marks they can ever ask in the HSC paper (from what I've seen) and sometimes they don't even ask. So don't worry too much about it, everyone is left with uncertainty when answering a question in an exam and perfecting this topic is not as easy as other topics. Practicing is probably the best way to increase your chance to get a question right as well as understanding and setting up a given situation.

2 marks means a lot to some people

 

[hit patel]

hmm... i am in year 11 and to become better at this all I did was expose myself to a range of question from past papers and books. This is not useless as stated somewhere above in the thread. It helps u recognise the kind of problems and what the examiners are asking for. your impulses become accustomed to that type of question if u do a few. If u dont believe it try it.

 

[braintic]

hmm... i am in year 11 and to become better at this all I did was expose myself to a range of question from past papers and books. This is not useless as stated somewhere above in the thread. It helps u recognise the kind of problems and what the examiners are asking for. your impulses become accustomed to that type of question if u do a few. If u dont believe it try it.

I don't think anyone stated that doing lots of questions is useless.

BTW, if you are seriously aiming for 99.95, I think those marks you are aiming for in each subject would see you get about 99.75

 

[hit patel]

I don't think anyone stated that doing lots of questions is useless.

BTW, if you are seriously aiming for 99.95, I think those marks you are aiming for in each subject would see you get about 99.75

Uhh... no..... I Used HSC 2013 Atar Calculator?

http://www.atarcalculator.com.au/