**OFFICIAL S1 OCR (Non-MEI) Thread 6th June 2014**

Can someone help with understanding Q 6 (i)(b) From Jan 09 paper please?

Q A test consists of 4 Algebra questions A B C D and 4 geometry questions G H I J. Find the probability that no two Geometry questions are next to each other.

To work out the no of permutations, the eg in the book (and Examsolutions video: http://www.examsolutions.net/maths-revision/statistics/permutations-combinations/permutations/tutorial-6.php) suggets:

1. Separate unrestricted items by spaces and put an extra space at start and end;

so : _ A _ A _ A _ A _ [ so that's 4 As and 5 spaces]
now,

there are 4! ways for A questions = 4!
There are 5 ways to put first G, 4 for second, 3 for 3rd and 2 for 4th.

we could also have: _ G _ G _ G _ G _

So, total no of ways = 4!x5x4x3x2 X2 [although the book example doesn't seem to x2]

But the answer (for number of permutations) is 4!x4!x2

I can see how they get it, but then its incompatible with the method in the book and in the video? What am I missing?

Do we need to know about venn diagrams?

I've done all the papers since 2005,, anyone know where I can find more questions which are related to OCR Topics?

Do we need to know about venn diagrams?

Can someone explain how rs(spearsman) could equal one and r (product moment coefficient) doesn't.

and the reverse

Imagine a line that curves upwards like an exponential graph. Or also imagine a line that kind of curves upwards and levels off. Here r does not equal 1 as the points are not in a straight line, but rs does equal 1 as y increases with x.

The reverse is not possible. If the points were to form an exactly straight line then the y ranks and the x ranks would have to increase with each other, so rs would also equal 1.

Thanks. Could you also explain how there can be two lines in the method of least squares. I get the idea of minimising the difference^2 between points, but not about the vertical or horizontal nature of a line that can fit between them.