Permutation question (math) [ ]

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The discussion revolves around solving a permutation problem using combinatorial methods. The user initially attempted to apply the formula 10!/(4!6!) but struggled with the concept of calculating paths to vertices in a grid-like structure. The solution involves understanding that the number of ways to reach a vertex is the sum of the ways to reach the preceding vertices, akin to Pascal's Triangle, but adapted for the specific problem context. The final answer is confirmed to be option 'c'.

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  • Understanding of permutations and combinations, specifically the formula for combinations.
  • Familiarity with Pascal's Triangle and its application in combinatorial problems.
  • Basic knowledge of grid-based pathfinding and vertex connectivity.
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permutation question (math) [urgent]

image here

please solve this problem for me. the correct answer is c.

i put 10!/(4!6!), then i know i am suppose to divide/subtract something, but i don't know what. (i have never done this kind of problem before.)
 
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hell I am stumped...
 
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:
1. 1. 1. 1. 1.  .  .
1. 2. 3. 4. 5.  .  .
1. 3. 6.        .  .
1. 4.10.  .  .  .  .
 .  .  .  .  .  .  .

See what I'm doing and why?
 
somebody please help
 
but how u going to 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...
 
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:
1. 1. 1. 1. 1.  .  .
1. 2. 3. 4. 5. 6.  .
1. 3. 6.       6.  .
1. 4.10.10.  .  .  .
 .  .  .  .  .  .  .
 
Im guessing the park is considered as one big square.
 
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.
 
why u have a 10 after the 10? how the get the second 10?
 
  • #10
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
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 a lot!
 

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