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by N:Dure » Thu Dec 23, 2010 6:18 pm
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by anshumishra » Thu Dec 23, 2010 6:28 pm
1 -> R
2,4,6,8 -> C, M
5 -> M
3, 7 -> V,P, D

So, total no. of arrangements = 1*2*3*2*1*2*3*2 = 144
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by Rahul@gurome » Thu Dec 23, 2010 6:51 pm
1st place – Monet – 1 way
2nd, 4th, 6th, 8th place – Cezzane or Monet – 2^4 -16 ways
5th place – Manet – 1 way.
2 - 3rd and 7th places are now remaining.
They can have Van Gogh, Pissaro or Degas in 3^2 = 9 ways.
So required number of ways of arranging the paintings is 1*16*1*9= 144 ways.
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by N:Dure » Thu Dec 23, 2010 8:53 pm
Thanks Anush & Rahul

@ Rahul the way you do it for C & M is more of counting right? Is this the only way to do it or u can do it in a comb way?
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by Rahul@gurome » Thu Dec 23, 2010 9:08 pm
N:Dure wrote:Thanks Anush & Rahul

@ Rahul the way you do it for C & M is more of counting right? Is this the only way to do it or u can do it in a comb way?
Yes. What you need to do is to find out that for each place, how many ways of putting a painting is possible.
Then multiply them all together to get the required answer.
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by N:Dure » Sat Jan 15, 2011 1:11 pm
Rahul@gurome wrote:
Yes.
Thanks Rahul
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