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by beatthegmatinsept » Thu Sep 23, 2010 8:01 am
IMO A.
This is like problem # 190 or 191 in OG 12, where they ask you to find the shortest possible length traveling in going 3 blocks north and 2 blocks east. NNNEE and so on.
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by goyalsau » Thu Sep 23, 2010 8:08 am
I am not done with OG 12 yet, so don't know much about that,
well you answer is correct but how to do it?
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by lokesh r » Thu Sep 23, 2010 8:47 am
(5!)/(3! x 2!)=10

IMO A.
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by beatthegmatinsept » Thu Sep 23, 2010 9:06 am
lokesh r wrote:(5!)/(3! x 2!)=10

IMO A.
The above is a quick way to do it.. I solved by considering diff combinations:
RRRBB
RRBBR
RBBRR
BBRRR
RBRRB
RBRBR
BRBRR
BRRBR
BRRRB
BRRRB
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by diebeatsthegmat » Thu Sep 23, 2010 10:20 am
goyalsau wrote:HI! guys.
3 red +2 blue= 5
thus 5!/2!3!=10
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by tlt2372 » Thu Sep 23, 2010 6:36 pm
This is a combinatorics problem - MGmat goes over this in detail in the Word Translations Strategy Guide.

5!/3!2! = 10
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