#195 from OG - Not sure

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#195 from OG - Not sure

by AleksandrM » Sat Apr 19, 2008 1:25 pm
So I came across this problem in the OG, and, treating it as an exam, just guessed. I guessed correctly but then I wanted to see the solution, which shows you the way you should absolutely NOT solve the problem. Using part of the GMAC approach, I solved it as I show below. I would like to know if I am right. I did get the right answer using the method.

Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above [the map is attached, that is the best I could do]. How many routes from X to Y can Pat take that have the minimum possible length?

A) 6
B) 8
C) 10
D) 14
E) 16

Using the map, after about three tries, you notice that the shortest way takes five steps. Therefore, how many ways can you move up and over (two different movement types) on the map using five steps. Therefore, I used the formula:

n!/k!(n - k)! = 5!/2!(5 - 2)!

5!/2 x 3! = 5 x 4/2 = 20/2 = 10

Let me know if this is correct. Thanks.
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by netigen » Sat Apr 19, 2008 4:07 pm
I think your reasoning is correct but the usage of the combination formula may not be correct approach in this case.

Since one has to take total of 5 steps and each step can be take in two different way, the total possibilities will be 5 x 2 = 10

5C2 is the number of ways to choose 2 things out of 5 which is not what we are doing here.

You should test by changing the number of block grids in the Q to find out if you reasoning and approach holds in all cases.

May be an expert can throw more light.

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by jasonc » Sun Apr 20, 2008 12:10 am
Using the combination formula n!/(n-k)! will not give you the right answer most of the time (in this case it did, but thats purely coincidental).

To both neti & alek's thought process - you guys considered that there are 2 different movements possible, and thats correct. However, you didn't take into account the fact that you're limited by 2 right moves & 3 up moves, so each step won't actually be 2 choices.

SolutionThis is a permutation problem with identical elements.

To get from x to y you need 5 moves total. 3 up moves & 2 right moves.
To find the ways of doing this, use the formula:
n!/a!/b!...etc where n is the total number of elements and a, b, c.. are the number of identical elements in distinct groups a, b,c.

So in this case we have 2 groups: up & right
we have 2 elements in the right group and 3 in the up group.

5!/2!/3! = 5*4/2 = 10.

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by mandy12 » Sun Apr 20, 2008 5:13 am
I found this formula somewhere for finding the number of shortest paths across a grid of m*n

(m+n-2)C(m-1) or (m+n-2)C(n-1)

In ur case m = 4 n =3 ....so we get 5C2