shobhitb wrote:
giddi77 wrote:
It is B! Made a mistake..
5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252
Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25
Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252
Required probability = 1-25/252 = 227/252
Giddi - Please elaborate on the logic behind the lines in bold. Thanks!
If you take a simple 2 by 2 path (see picture), From X to Y the shortest path will always have only 4 LEGS (by a LEG I mean either a horizontal or vertical bar) on the picture.
If you denote R = right , D = down, then one of the shortest paths form X to Y could be
R-R-D-D
Similarly, D-D-R-R, D-R-R-D, D-R-D-R, R-D-R-D, R-D-D-R are also shortest possible unique routes. In all there are 6 routes.
Rather than counting all of them we can say that, all possible shortest routes are unique arrangements of 4 letters R, R, D, D
Hence number of routes = 4!/(2!*2!) = 6
In this problem we have 5 Right and 5 Down LEGS.. Hence total routes = 10!/(5!*5!)
Similarly from A->C we have 1-D, and 4-Rs.. Hence total routes from A-> C = 5!/4!
From C-> we again have 1-R and 4-Ds. Hence 5!/4! routes.
From A->B->C = (5!/4!)*(5!/4!)
HTH! BTW when I first got similar problem in my PP test, I couldn't solve it either. I tried to solve this problem for next 2 days and then gave up
Found the solution in
OG.
Attachments
Shortest_route.GIF [ 1.26 KiB | Viewed 2032 times ]
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