a city has p parellel roads running east west and Q parellel reads running north south. how many rectangles are formed with their sides along these roads if the distance bet every consequtive parellel road is the same. how many shortest possible routes are there to go from one corner to the diagonally opposite corner.

 

ans:  (p+q -2)! / ((p-1) (q-1) !)

quest. is confusing...

 

but see that we can compress each road into a line and thus no. of rectangles will be pq...

how is the ans. gven this?

compressing roads into lines each gaps bcom rectangles formed and also each chowk bcom intersection point of lines....

 

shortest possible routes..

watever u do u will have to cover the boder line much dist. at min....

do it in rectangle lik a chess board...go through edges anyway frm one end to other

 progress frm one end say bottom left most... cornor... going along line (compreesed roads) go either up or right and reach other cornor.. cant go down and left... as shortest..( all will yield same dist.. include thm all usin perm. comb.)

 

 

hope this helps

Cheers!!!!!!!

 

 

no. of rectangles is pC2 * qC2 for obvious reasons

 

lets consider a right as the letter R and up as the letter U

Now for a person to move from one diagonal to the other, he has to move thro p-1 rights and q-1 ups i.e. its eqvt to arranging a sequence of p-1 R's and q-1 U's

So no. of ways is ((p-1)+(q-1))!/(p-1)!(q-1)!