# Homework Help: Permutation question (math) [ ]

1. Apr 25, 2005

### F|234K

permutation question (math) [urgent]

image here

i put 10!/(4!6!), then i know i am suppose to divide/subtract something, but i dont know what. (i have never done this kind of problem before.)

2. Apr 25, 2005

### JamesU

hell im stumped...

3. Apr 25, 2005

### BicycleTree

You can just add up the ways to get to each vertex. Here are the first few (period represents a vertex, number before each vertex is the number of ways to get to it):
Code (Text):

1. 1. 1. 1. 1.  .  .
1. 2. 3. 4. 5.  .  .
1. 3. 6.        .  .
1. 4.10.  .  .  .  .
.  .  .  .  .  .  .

See what I'm doing and why?

4. Apr 25, 2005

### F|234K

somebody plz help

5. Apr 25, 2005

### F|234K

but how u gonna get the numbers for the dots in the park!

i did try to solvei t using pascal triangle thing...but i got stuck on the park thing...

6. Apr 25, 2005

### BicycleTree

There are no dots in the park. And it's not exactly Pascal's triangle because of the park. Here's a hint (some more vertexes filled in):
Code (Text):

1. 1. 1. 1. 1.  .  .
1. 2. 3. 4. 5. 6.  .
1. 3. 6.       6.  .
1. 4.10.10.  .  .  .
.  .  .  .  .  .  .

7. Apr 25, 2005

### whozum

Im guessing the park is considered as one big square.

8. Apr 25, 2005

### BicycleTree

No, the park is considered a blank. The number of ways to get to any vertex is equal to the sum of the number of ways to get to any of the immediately preceding vertices.

9. Apr 25, 2005

### F|234K

why u have a 10 after the 10??? how the get the second 10?

10. Apr 25, 2005

### BicycleTree

For example take the first 10 you get, at vertex (4, 3) by row, column. At vertex (3, 3) you have a 6, and at vertex (4, 2) you have a 4. You can get to the 10 one of two ways: through (3, 3) by going south, and through (4, 2) by going east. There are 6 ways to get to (3, 3) so there are 6 ways to get to (3, 3) and then go south. There are 4 ways to get to (4, 2) so there are 4 ways to get to (4, 2) and then go east. So you have 4 ways + 6 ways = 10 ways for vertex (4, 3).

If a vertex only has one other vertex leading into it--say the other vertex has 7 ways to get to it--then how many ways can you get to that second vertex?

11. Apr 25, 2005

### F|234K

oh....i think i got it....thanks for this "If a vertex only has one other vertex leading into it--say the other vertex has 7 ways to get to it--then how many ways can you get to that second vertex?"

again..thanks alot!!!!