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The diagram below shows the various paths along which a [#permalink]

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25 Apr 2012, 07:54

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82% (01:25) correct
18% (00:28) wrong based on 77 sessions

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The diagram below shows the various paths along which a mouse can travel from point X, where it is released, to point Y, where it is rewarded with a food pallet. How many different paths from X to Y can the mouse take if it goes directly from X to Y without retracting any point along a path?

The diagram below shows the various paths along which a mouse can travel from point X, where it is released, to point Y, where it is rewarded with a food pallet. How many different paths from X to Y can the mouse take if it goes directly from X to Y without retracting any point along a path?

A. 6 B. 7 C. 12 D. 14 E. 17

I know my approach is wrong. Can you please help me understand why the permutation formula does not work for this problem? ( We are counting without replacement and the order matters, so I tried n!/(n-3)! where n=7

Thank you, all!

It's not clear what's you logic behind applying permutation to this problem. I guess 3 is # of the forks on the road, but we are not choosing those 3 out of 7 (?).

Anyway, the problem is about simple counting. There are 3 forks in the road: 2 choices for the first one, 2 for the second and 3 for the third. Hence total # of ways is 2*2*3=12.

Re: The diagram below shows the various paths along which a [#permalink]

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25 Apr 2012, 08:25

Bunuel wrote:

It's not clear what's you logic behind applying permutation to this problem. I guess 3 is # of the forks on the road, but we are not choosing those 3 out of 7 (?).

Anyway, the problem is about simple counting. There are 3 forks in the road: 2 choices for the first one, 2 for the second and 3 for the third. Hence total # of ways is 2*2*3=12.

Answer: C.

7 paths to chose from and you must pick 3 to get to the destination..

I haven't seen a permutation problem yet in the OG, so maybe the difference in identifying when to use it will show when I see one.

It's not clear what's you logic behind applying permutation to this problem. I guess 3 is # of the forks on the road, but we are not choosing those 3 out of 7 (?).

Anyway, the problem is about simple counting. There are 3 forks in the road: 2 choices for the first one, 2 for the second and 3 for the third. Hence total # of ways is 2*2*3=12.

Answer: C.

7 paths to chose from and you must pick 3 to get to the destination..

I haven't seen a permutation problem yet in the OG, so maybe the difference in identifying when to use it will show when I see one.

As you can see from the solution there are 12 different paths not 7. There are 2+2+3=7 different line segments separated by 3 forks.

Re: The diagram below shows the various paths along which a [#permalink]

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30 Aug 2013, 09:03

Bunuel wrote:

emil3m wrote:

Bunuel wrote:

It's not clear what's you logic behind applying permutation to this problem. I guess 3 is # of the forks on the road, but we are not choosing those 3 out of 7 (?).

Anyway, the problem is about simple counting. There are 3 forks in the road: 2 choices for the first one, 2 for the second and 3 for the third. Hence total # of ways is 2*2*3=12.

Answer: C.

Buneul, Could you please explain why the answer is 12 and not 7. Since there are 7 separate forks, how could there be twelve separate ways the mouse could travel? Why are you multiplying the numbers instead of adding?

It's not clear what's you logic behind applying permutation to this problem. I guess 3 is # of the forks on the road, but we are not choosing those 3 out of 7 (?).

Anyway, the problem is about simple counting. There are 3 forks in the road: 2 choices for the first one, 2 for the second and 3 for the third. Hence total # of ways is 2*2*3=12.

Answer: C.

Buneul, Could you please explain why the answer is 12 and not 7. Since there are 7 separate forks, how could there be twelve separate ways the mouse could travel? Why are you multiplying the numbers instead of adding?

Thanks!

There are 3 forsk not 7:

As for multiplication: it's called Principle of Multiplication. If one event can occur in m ways and a second can occur independently of the first in n ways, then the two events can occur in m*n ways.

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