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I tried to just draw it out, but with the time limit I end up just confusing myself. Is there a more simple way?
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30 Jun 2008, 16:30
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this is a combinatorix problem, so the easiest way is to classify the choice the person can make at each intersection. In this example, he can go left twice and down three times, and has to make a total of five decisions. Since there is no difference between lefts (conceptually, one left is the same as another), you want to keep the lefts and the downs in the same group. then, to build the equation, you write out the number of decisions (factorial) for the numerator and the groups (factorial) for the denominator, as follows:
1 2 3 4 5
------------
L L D D D
The equation will then look like this:
5!
-----
2! 3!
or
4*5
----
2
=
10
This is because there are two groups on the bottom of 2 possibilities and 3 possibilities, respectively. So you just divide the total number of choices by the number of constraints.
1 2 3 4 5
------------
L L D D D
The equation will then look like this:
5!
-----
2! 3!
or
4*5
----
2
=
10
This is because there are two groups on the bottom of 2 possibilities and 3 possibilities, respectively. So you just divide the total number of choices by the number of constraints.
30 Jun 2008, 16:54
Kudos
Bookmarks
i agree with the answer C. Shortest path consists of 5 blocks with 2 blocks going right.
C(5,2)=5*4/2 = 10 -> C
C(5,2)=5*4/2 = 10 -> C
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
