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 Posted: Wed Jul 03, 2013 3:06 pm

Joined: Wed Sep 12, 2012 7:02 pm
Posts: 24
Could someone Please help with the following question taken from The North London Consortium 2010 Group 2 paper. It is question number 41.

http://www.clsg.org.uk/_files/EC8962FF7FD78D3102EDCC92E030F7DF.pdf

I've added a link to the paper

Thanks

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 Posted: Fri Jul 12, 2013 1:05 pm

Joined: Thu Jan 03, 2008 8:26 am
Posts: 1326
Location: Watford, Herts
Part (a): You need to shade one square below the diagonal and one above, but once you've chosen the first square the position of the second is fixed. There are 6 squares below the diagonal, so 6 possibilities.

Part (b): Since the two squares must be on the diagonal, the question is how many different ways can you choose 2 things from 4. There are 4 ways of choosing the first, and for each of those there are 3 possibilities for the second (4*3 = 12), but we've counted each pair twice, so the answer is 4*3/2 = 6. (look up permutations and combinations)

Remaining part: This can only work if both squares are on a diagonal, and again once you've shaded one square you have no choice for the other one. There are 4 diagonal squares in the bottom half of the big square, so those are the only possibilities:
Code:
*---  ----  ----  ---*
----  -*--  --*-  ----
----  --*-  -*--  ----
---*  ----  ----  *---

The last one could be done just by trying all the possibilities and making sure not to repeat any.

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 Posted: Fri Jul 12, 2013 5:38 pm

Joined: Wed Sep 12, 2012 7:02 pm
Posts: 24
Thank You so much WP for the answer

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