is there a really fast way of solving PS problem #195 from OG 11ed ?
pat walks from intersection X to intersection Y problem ?
Thnx,
-Badri
og 11ed problem
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- givemeanid
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Yeah. And do not even get me started on ones who post screen shots for problems with plain text...
So It Goes
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.
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.
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!
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!
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.
(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.