Unformatted text preview: by 7! 4! · 3! = 35 . Problem 22 (paths on a grid through a speciFc point) One can think of the problem of going through a speciFc point (say P ) as counting the number of paths from the start A to P and then counting the number of paths from P to the end B . To go from A to P (where P occupies the (2 , 2) position in our grid) we are looking for the number of possible unique arraignments of two “U”’s and two “R”’s, which is given by 4! 2! · 2! = 6 , possible paths. The number of paths from the point P to the point B is equivalent to the number of di²erent arraignments of two “R”’s and one “U” which is given by 3! 2! · 1! = 3 ....
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Penn State.
 Winter '08
 G.JOGESHBABU
 Counting, Probability

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