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# A Vstudy problem

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Intern
Joined: 04 Jul 2003
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25 Aug 2003, 05:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can anyone help me figure this problem ?

Thank you.
Manager
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25 Aug 2003, 05:49
minimum pat: 2 ups, 3 rights... UURRR...
now, in how many ways can we arrange UURRR? 5!/(3!*2!)=10
Manager
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25 Aug 2003, 05:58
javropu wrote:
minimum pat: 2 ups, 3 rights... UURRR...
now, in how many ways can we arrange UURRR? 5!/(3!*2!)=10

I agree with your answer, but just to clarify, the route is 3 ups and two rights.

UUURR = (5!)/(3!*2!) = 10
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Sept 3rd

Senior Manager
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02 Jul 2005, 17:28
mciatto wrote:
javropu wrote:
minimum pat: 2 ups, 3 rights... UURRR...
now, in how many ways can we arrange UURRR? 5!/(3!*2!)=10

I agree with your answer, but just to clarify, the route is 3 ups and two rights.

UUURR = (5!)/(3!*2!) = 10

plz can anyone tell me from what formula you got that (5!)/(3!*2!) =
and why plz show full explanation
thanks

regards mandy
Director
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03 Jul 2005, 00:29
the formula is

(total # of squares)!/((# of squares on one side)!(number of squares on the other side)!)
Senior Manager
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03 Jul 2005, 13:06
sparky wrote:
the formula is

(total # of squares)!/((# of squares on one side)!(number of squares on the other side)!)

Thanks Sparky just one thing please the total number of square is 6 But the previous posts had 5
my first instinct was to think about permutations

regards

mandy
03 Jul 2005, 13:06
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# A Vstudy problem

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