Posted by **Jane** on Thursday, October 27, 2011 at 6:50pm.

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: (n-r)! / (n-r+1)!

3) solve (n+2)!/ (n-1)! = 210

- Data management -
**MathMate**, Thursday, October 27, 2011 at 8:59pm
A),B),C)

It's like finding the number of ways to arrange 4 1's and 7-0's.

The number of ways is given by (m+n)!/(m!n!)

Experiment first with a grid of 2x2, and then a 2x3.

2.

(n-r)!/(n-r+1)!

=(1.2.3...n-r)/(1.2.3...(n-r).(n-r+1))

=1/(n-r+1)

3.

similarly,

(n+2)!/(n-1)! = 210

(1.2.3...(n+2))/(1.2.3...(n-1)) = 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

- Data management -
**Anonymous**, Thursday, October 27, 2011 at 9:19pm
thank even though im not sure if u woule see this

- Data management -
**MathMate**, Thursday, October 27, 2011 at 10:47pm
You're welcome!

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