I agree, permutation and combinations is hard! Even at uni I did a subject called Combinatorial Probability and it was one of the toughest subjects of the whole Actuarial Studies degree! I also agree that nothing beats practicing the broadest range of questions possible. These questions don't really have a rigid formulaic approach to them, so the best you can do is hope that anything you encounter on an exam is something you've already been exposed to.Definitely one of the hardest topics and still at the forefront of the NESA test writers’ minds (as indicated by the last question of the 2020 Ext 1 HSC) despite being a Yr11 topic. What makes this topic so hard is the very fact that you can be asked anything. However, when doing it you begin to find that a lot of the questions are different variations of the same thing. So my advice to you would be to make a note of how to solve these standard question types and how to solve them. After this, look at as many niche questions as possible (in past papers) and try to solve them yourself then look at the marking guideline’s approach. There are some specialised techniques you can make a note of in these questions. As a general tip I would try to break down a question into simpler cases which you can then solve for. Hope this helps!
Whilst I agree there is a lot of variation most of perms and combs is formulaic just like the rest of the syllabus (by most I mean like 80% whereas 20% is unknown or has a trick). You will find a lot of questions use the same trick whereas a few require you to try something new.I am about to start this topic, so i was wondering if there was any tips, tricks, or anything in general to lookout for
Thats true, as i do alot of questions i see that there are alot of variationWhilst I agree there is a lot of variation most of perms and combs is formulaic just like the rest of the syllabus (by most I mean like 80% whereas 20% is unknown or has a trick). You will find a lot of questions use the same trick whereas a few require you to try something new.
Would an example being using tree diagram instead of multiplication and addition principalso only do the long way if you have the time and you are confident you can get the answer.
That is so correct, as i do the questions, i can see that there is very little change, although, how do i counter the proof questions, those tend to be harderHowever, when doing it you begin to find that a lot of the questions are different variations of the same thing. So my advice to you would be to make a note of these standard question types and how to solve them.
Give an example.That is so correct, as i do the questions, i can see that there is very little change, although, how do i counter the proof questions, those tend to be harder
What do you mean in particular? Could you provide an example?That is so correct, as i do the questions, i can see that there is very little change, although, how do i counter the proof questions, those tend to be harder
I thought they were essentially the same process. When you draw a tree diagram, you multiple the probabilities of consecutive events along each branch, then you add the final products of the branches that fit the stated conditions/outcomes.Would an example being using tree diagram instead of multiplication and addition principal
I was thinking of a question like:Would an example being using tree diagram instead of multiplication and addition principal
Well the “S” can occur either before or after the “H” making the probability 1/2.I was thinking of a question like:
Consider the "words" that can be made by rearranging the letters of MATHEMATICS. How many different words can be made in which the "S" occurs before the "H"?
Sorry misread! It’s not a probability question. Oops!Well the “S” can occur either before or after the “H” making the probability 1/2.
Haha I have made that mistake a few times!Sorry misread! It’s not a probability question. Oops!
Correct.Sorry misread! It’s not a probability question. Oops!
Then it becomes 11!/(2x2!x2!x2!).
How did you get this answer?, the only way i can think off getting this is by listing out all the options for S behind H. I am super confused rn11!/(2x2!x2!x2!).
Half the total ways are when S is to the left of H and half are when S is to the right of A. So simply work out the total number of ways of arranging the letters in general then divide by 2 (what Bianca did).How did you get this answer?, the only way i can think off getting this is by listing out all the options for S behind H. I am super confused rn