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# The diagram below shows the various paths along which a

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Intern
Joined: 24 Apr 2012
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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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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-diagram-above-shows-the-various-paths-along-which-a-mous-144271.html
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 129061 [2], given: 12190

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

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25 Apr 2012, 08:15
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Expert's post
emil3m wrote:
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.

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Kudos [?]: 129061 [2], given: 12190

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Joined: 24 Apr 2012
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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.

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.

Kudos [?]: 1 [0], given: 2

Math Expert
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Posts: 41891

Kudos [?]: 129061 [1], given: 12190

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

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25 Apr 2012, 08:31
1
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Expert's post
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.

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.

As for permutation. Try Combinatorics chapter of Math Book for theory: math-combinatorics-87345.html

Also try some questions on combinations to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52
Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html (there are some about permutation too)

Hope it helps.
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Manager
Joined: 28 Jan 2013
Posts: 90

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Location: United States
Concentration: Social Entrepreneurship, General Management
GMAT 1: 650 Q44 V35
GPA: 3.1
WE: Education (Education)
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.

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!

Kudos [?]: 19 [0], given: 42

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 129061 [1], given: 12190

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

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31 Aug 2013, 05:57
1
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Expert's post
Bunuel wrote:
emil3m 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.

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-diagram-above-shows-the-various-paths-along-which-a-mous-144271.html
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Kudos [?]: 129061 [1], given: 12190

Re: The diagram below shows the various paths along which a   [#permalink] 31 Aug 2013, 05:57
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