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og 11ed problem

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by beny » Tue Aug 14, 2007 12:06 am
No one is going to look up the question to help you. If you want help, at least be active enough to type out the question.
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by gabriel » Tue Aug 14, 2007 12:41 am
..Seriously dude .. the least u can do is type out the question .. plz do not expect people to go and fetch the question for u and then respond to your queries ..
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by givemeanid » Tue Aug 14, 2007 5:26 am
Yeah. And do not even get me started on ones who post screen shots for problems with plain text...
So It Goes
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reg: problem

by badri077 » Tue Aug 14, 2007 8:22 am
sorry folks. the problem had a figure which I couldn't draw very easily. Hence the pointer to the problem.
Will write up the whole problem and try to draw the figure once I get home tonight.

-B
P.S. From what I remember-
imagine a 1st quadrant grid. How many different shortest paths are possible from (0,0) to (3,4) travelling only along the grid lines (going from y=0 to y=4, x=0 to x=3
I'll confirm this evening.
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by beny » Tue Aug 14, 2007 12:19 pm
I remember this question... I believe the answer was 10?

By taking the shortest route possible, he can only travel Up (U) three or Right (R) two.

The combinations are:

UUURR
UURRU
URRUU
RRUUU
UURUR
URURU
RURUU
URUUR
RUURU
RUUUR

In other words:

5!/3!*2!
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by 800GMAT » Thu Aug 16, 2007 3:31 pm
here..lemme post the question
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by 800GMAT » Thu Aug 16, 2007 3:36 pm
beny, one question----what does 5! denote...........
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by beny » Thu Aug 16, 2007 5:46 pm
The combinatoric way to solve this question is the "anagram approach". In other words, how many words can you make with a set of letters via rearranging the letters. The way to figure this out is:

(Total number of letters)! / (Number of each repeating letter)!

Here, the total number of letters is 5, there are 2 repeating R's and 3 repeating U's.
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by 800GMAT » Fri Aug 17, 2007 8:03 am
got it.....thnkx
:)
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