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# A Vstudy problem

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Intern
Joined: 04 Jul 2003
Posts: 7
Location: Chennai
Followers: 0

Kudos [?]: 5 [0], given: 0

A Vstudy problem [#permalink]  25 Aug 2003, 05:44
Can anyone help me figure this problem ?

Thank you.
Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

Kudos [?]: 0 [0], given: 0

minimum pat: 2 ups, 3 rights... UURRR...
now, in how many ways can we arrange UURRR? 5!/(3!*2!)=10
Manager
Joined: 10 Jun 2003
Posts: 210
Location: Maryland
Followers: 2

Kudos [?]: 4 [0], given: 0

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
_________________

Sept 3rd

Senior Manager
Joined: 30 May 2005
Posts: 278
Followers: 1

Kudos [?]: 10 [0], given: 0

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
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

the formula is

(total # of squares)!/((# of squares on one side)!(number of squares on the other side)!)
Senior Manager
Joined: 30 May 2005
Posts: 278
Followers: 1

Kudos [?]: 10 [0], given: 0

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
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