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CasperMonday
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MGMAT: Bill and cards 29 Aug 2009, 03:49
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0% (00:00) correct 100% (03:08) wrong based on 1 sessions

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Guys, the highlighted answer is not the OA. It is just my guess. I don't know the OA.




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MGMAT: Bill and cards 29 Aug 2009, 05:26
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an interesting question.
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CasperMonday
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MGMAT: Bill and cards 29 Aug 2009, 06:12
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Alright, here is my solution

\(1-\frac{1C12*1C10*1C8*1C6}{4P12}=17/33\)

But what I can't really understand is why I divide by 4P12 and not, for example by 4C12?
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MGMAT: Bill and cards 29 Aug 2009, 07:51
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CasperMonday
Alright, here is my solution

\(1-\frac{1C12*1C10*1C8*1C6}{4P12}=17/33\)

But what I can't really understand is why I divide by 4P12 and not, for example by 4C12?
Show more

I have not calculated the value of the equation which you have given, but we cannot apply permutation here because the order of cards does not matter. sequence 1,2,1,4 is the same as 1,1,2,4, we just need to check the pairs ( not the sequence of cards ).

Answer is 17/33.
4 cards from different pairs can be selected in 6C4*2*2*2*2 ways.
a sequence (without order being important) of 4 cards can be selected in 12C4 ways.

Req ans will be 1 - (6C4*2*2*2*2/12C4) = 17/33.



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Hi there,
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Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
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Thank you for understanding, and happy exploring!
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