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.
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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