# Thread: word problem

1. ## word problem

Ok this is kind of tough to explain. It's an EC problem but i have no idea where to start, should I find a pattern?

There is a checkerboard that is 60 squares in length by 58 squares in width.

You want to travel from the top left corner to the bottem right corner. But you can only move down or right.

How many paths can you take?

Heres an example of a possible path if the board was smaller and perfectly square. As you can see you can only move right or down.

2. ## A hint

1 1 1 1
1 2 3 4
1 3 6 10

3. This is a classic problem and solved many times in the forum.

http://www.mathhelpforum.com/math-he...m-solving.html
http://www.mathhelpforum.com/math-he...ing-paths.html

4. ok the pascals triangle thing helps me, but how can I solve it for 58 by 60 squares without writing everything out?

I got this off of wikipedia but i dont know if it is relevant or where to plug the numbers in

""This construction is related to the binomial coefficients by Pascal's rule, which states that if
is the kth binomial coefficient in the binomial expansion of (x + y)n, where n! is the factorial of n, then
for any nonnegative integer n and any integer k between 0 and n.[1]""

5. Hint: Pascal's triangle is a list of combinations.

Code:
          1

1     1

1     2     1

1     3     3     1
Code:
                C(0,0)

C(1,0)    C(1,1)

C(2,0)   C(2,1)    C(2,2)

C(3,0)    C(3,1)    C(3,2)   C(3,3)

6. Think about simply. You must move down 60 places and to the right 58 places.
How many ways can one arrange 60 Ds and 58 Rs in a line.
Answer: $\frac {118!} {(60!)(58!)}$.

7. oooh I get it now, doing it on a checkerboard is just the same as arangeing the letters D, R in a line so it's
118!

8. Originally Posted by stoked365
oooh I get it now, doing it on a checkerboard is just the same as arangeing the letters D, R in a line so it's
118!
No, the answer is $\frac {118!}{(60!)(58!)}$ because we have repeated items.

How many ways can the letters is the word MISSISSIPPI be arranged?
Answer: $\frac {11!} {(2!)(4!)(4!)}$ because we must account for the identical letters that are repeated.

9. i get it 100% now, thanks for the help