Data management
posted by Jane .
1) consider the number of grid paths from the orgin in a coordinate graph to each of the following points:
a) (4,7) b) (3,7) c) (4,6)
2) simplify: (nr)! / (nr+1)!
3) solve (n+2)!/ (n1)! = 210

A),B),C)
It's like finding the number of ways to arrange 4 1's and 70's.
The number of ways is given by (m+n)!/(m!n!)
Experiment first with a grid of 2x2, and then a 2x3.
2.
(nr)!/(nr+1)!
=(1.2.3...nr)/(1.2.3...(nr).(nr+1))
=1/(nr+1)
3.
similarly,
(n+2)!/(n1)! = 210
(1.2.3...(n+2))/(1.2.3...(n1)) = 210
n.(n+1).(n+2) = 210
Start with cube root of 210 = 5.9...
so try 5.6.7=210 OK.
So n=5,n+1=6,n+2=7 
thank even though im not sure if u woule see this

You're welcome!