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.
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