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funny thing is, the solutions are web-wide easy to find, yet the papers are hidden?? what logic is that?

So does anyone have it?

I would like to ask the same question

yes, 540 or 1080. Personally i think it is 1080 because urns are non-identical

what do you mean? What i wrote there was trying to clarify whether to divide 2! Or not

so for example, Case 1: balls B,C,D,E in urn A, ball A in urn B and ball F in urn C Case 2: balls B,C,D,E in urn A, ball F in urn B and ball A in urn C Would you call that same outcome or different? I would say, they are same outcome IF urns are identical, but this is clearly not the case...

no, but if you think about it, ball A goes into urn B or ball A goes into urn C is 2 different outcomes since the urns are NON-IDENTICAL, correct?

yes, i did, searched mixtures of words "2012 extension 2 HSC paper", etc and i can find solution but not the actual PAPER

My way of thinking is there are 3 cases: Case 1: one urn has 4 balls, one urn has 1, another urn has 1. So we do 6C4 * 2C1 * 1C1 to choose the balls, then arrange them so 3!, and we DO NOT divide by 2! because the urns are NOT IDENTICAL, is this correct? This is the part that i am confused, to...

Does anyone know where i can download the paper?

Q: find the number of ways that 6 different coloured balls can be placed in 3 non-identical urns so that no urn is empty. Answer 1: 540 Answer 2: 1080 Which one is correct and why?

bump

kings trial 2012 anyone willing to share?

He probably has a friend/relative done the paper but has question he cant answer.

Can someone please upload sydney grammar and/or kings 2012 trial paper!!! I reckon they are hard and i would like to practise on that asap.

If you can upload the paper first, then i can make one for you.

mathman help?

I still don't get why the jars with 2 marbles are identical? aren't the jars different and so are the marbles?

once you chose 52C1 * 3C1 /2!, you have a pair, and you DONT want another pair from then on. you have 48 cards to choose from (since you dont want any of the first 2 cards and the 2 cards that has the same number), so 48C1, the next one cant be the same number, so choose one from 44, ie 44C1...

Question: How many ways for 5 DIFFERENT marbles to go in 3 jars so that 2 jars contains 2 marbles and the last jar contains one marble? Solution: 5C2 * 3C2 * 1C1 * 3! / 2! What my problem is: i understand that we firstly choose 2 out of 5, then 2 out of the remaining 3, then 1 out of 1, then...