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I wouldn't have a clue :p. If I could get 95+ in 4U then I would be more than happy.

Something around 80-90 out of 120 would leave me contented.

No problem :)

First of all there are three options where 2 rooms are empty. You can put all the people in room 1, all the people in room 2 or all the people in room 3. This gives you '3' rather than 1⁵. If only one room is to be empty you are left with two rooms to distribute the people between...

i) For you first question I'm sure you will agree that there are 2ⁿ ways to arrange n people between two rooms right? What you need to realise is that in one of these situations *everyone* has been put into room 1 and that in another situation *everyone* has been put into room 2...

If there's any theory I geuss it's that the number of ways to arrange p objects amongst q rooms (where each object and room is distinct) is q^p. The includes options where rooms are empty. The logic of this can be understood pretty easily from the example in i) where you have n...

Starting with part i) the number of ways to place n students into 2 rooms so that neither room is empty = (2ⁿ - 2) since there are 2 options for each student and you have to subtract the two cases where the first or the second room is empty. For part ii) it may well be helpful to...

Nope :p... well, maybe, but I don't think they expect them to be right.

That seems pretty similar to what I'm trying to do. Where possible I'm trying to do Q1-4 in about an hour because you need to leave time for the harder stuff but you need enough time to earn marks where you can get them. I then give about 40 mins - 60 mins to Q5-6 and the rest of the time to 7...

Go the tigers :D. I live in the middle of balmain and you should hear the number of cheering drunks, it's hilarious.

The top pecentiles have a large range of possible scores. Technically someone could get that mark but they would have to be in the upper 100th percentile for every section.

I think the answer for that case is still 1/365. It seems to be the same question. The assumption you make in the original question is that "someone's birthday is on a given day in the year". The probability that another person of an unknown birthdate is also born then is 1/365. You would get...

Noone needs a formula, the general maths people have it right.

It was a pun made in jest. I geuss you don't get the joke?

Because I made the invalid assumption that A, B and C are in line with the mast :p. Go with brett's method, it's solid.

You can probably do this part in a similar fasion: As before there are 6! ways you can arrange the letters of NUMRAT in a line. You then get 7 spaces... ₁N₂U₃M₄R₅A₆T₇ In this case you need to choose 2 of 7...

I was just thinking about it and I may have hit upon the way they got their answer. Say you take the letters of NUMRAT. There are 6! ways you can arrange them in a line. If you arrange them in a line you get 7 spaces in between and on either side of the letters where you could place the E's...

One thing I thought I'd mention is that the permutations with 2 E's touching also includes the cases where three E's are touching so if you take away the (EE) situation and the (EEE) then you're taking away the same thing twice.

Ahh, there's the trick.

One thing you could do to make life easier is to take a factor of 8¹⁷ out: 8¹⁷[1 - ¹⁷C₁(5/8) + ¹⁷C₂(5/8)² - ¹⁷C₃(5/8)³ + ... - (5/8)¹⁷] If you kept (5/8) on the [ANS] -...