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A man travels to his home from his current location on the r
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15 Sep 2017, 07:45
15 Sep 2017, 07:45
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Question Stats:
37% (01:35) correct
62% (02:01) wrong based on 61 sessions
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A man travels to his home from his current location on the rectangular grid shown above. If he may choose to travel north or east at any corner, but may never travel south or west, how many different paths can the man take to get home?
(A) 12
(B) 24
(C) 32
(D) 35
(E) 64
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Re: A man travels to his home from his current location on the r
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24 Apr 2018, 16:01
24 Apr 2018, 16:01
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This is a hidden combinatorics problem. In order to get home, he must travel north 3 times, and east 4 times, in any particular order. So the real question is: how many ways can you arrange 3 North turns and 4 East turns?
Think of it this way: how many ways would there be to arrange 7 turns total? This should give us 7! arrangements. However, we need to account for the fact that the North turns are all interchangeable, as are the East turns. For example, if we think of the 3 North turns as N1, N2, and N3, then one arrangement within our 7! would be N1N2N3EEEE. Another would be N1N2N3EEEE. In fact, we would have 6 arrangements that all start with 3 North turns in a row. But these shouldn't count as separate routes. If we alternate East, then North, we'd get ENENENE, but again there are 6 ways we could be arranging the North turns, even though they're all really the same route. Thus we find that the 7! number of configurations should be divided by 3!, since that's how many ways you can arrange 3 identical North turns. Similarly, we should divide it again by 4!, since that's how many arrangements there are of 4 identical East turns.
Therefore our equation is:
7!/(3!4!)
or
(7x6x5x4x3x2x1)/(3x2x1x4x3x2x1)
which equals 35, so the answer is D.
Think of it this way: how many ways would there be to arrange 7 turns total? This should give us 7! arrangements. However, we need to account for the fact that the North turns are all interchangeable, as are the East turns. For example, if we think of the 3 North turns as N1, N2, and N3, then one arrangement within our 7! would be N1N2N3EEEE. Another would be N1N2N3EEEE. In fact, we would have 6 arrangements that all start with 3 North turns in a row. But these shouldn't count as separate routes. If we alternate East, then North, we'd get ENENENE, but again there are 6 ways we could be arranging the North turns, even though they're all really the same route. Thus we find that the 7! number of configurations should be divided by 3!, since that's how many ways you can arrange 3 identical North turns. Similarly, we should divide it again by 4!, since that's how many arrangements there are of 4 identical East turns.
Therefore our equation is:
7!/(3!4!)
or
(7x6x5x4x3x2x1)/(3x2x1x4x3x2x1)
which equals 35, so the answer is D.
Re: A man travels to his home from his current location on the r
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21 Jun 2021, 22:47
21 Jun 2021, 22:47
2
Given that the man can only move north and east, he must advance
exactly 7 blocks from his current location to get home regardless of which
path he takes. Of these 7 blocks, 4 must be moving east and 3 must be moving
north.
The problem can then be rephrased as follows: “Of the 7 steps, when does the
man choose to go east and when does he choose to go north?” Labeling each
step as N for north and E for east, you can see the problem as the number of
unique rearrangements of NNNEEEE (e.g., this arrangement corresponds to
going north 3 times and then east 4 times straight to home). This is
determined by 7!/(3!*4!) which is 35.
exactly 7 blocks from his current location to get home regardless of which
path he takes. Of these 7 blocks, 4 must be moving east and 3 must be moving
north.
The problem can then be rephrased as follows: “Of the 7 steps, when does the
man choose to go east and when does he choose to go north?” Labeling each
step as N for north and E for east, you can see the problem as the number of
unique rearrangements of NNNEEEE (e.g., this arrangement corresponds to
going north 3 times and then east 4 times straight to home). This is
determined by 7!/(3!*4!) which is 35.
