Every morning, Casey walks from her house to the bus stop. She always travels exactly nine blocks from her house to the bus, but she varies the route she takes every day. How many days can Casey walk from her house to the bus stop without repeating the same route?
The original problem includes the following drawing:
This is an ARRANGEMENT problem.
To travel to the bus, Casey must move down 5 times (DDDDD) and to the left 4 times (LLLL).
Any arrangement of the 9 letters DDDDDLLLL will be composed of 5 movements downward and 4 movements leftward.
Thus, any arrangement of DDDDDLLLL represents a possible route:
If Casey travels DDLLDDLDL, she will travel downward 5 blocks and leftward 4 blocks, a route that will bring her to the bus stop.
If Casey travels DLDLDLDLD, she will travel downward 5 blocks and leftward 4 blocks, a route that will bring her to the bus stop.
And so on.
Thus, the number of possible routes is equal to the number of ways to arrange DDDDDLLLL.
Number of ways to arrange 9 elements = 9!.
But when an arrangement includes IDENTICAL elements, we must DIVIDE by the number of ways to arrange the identical elements.
The reason: when the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 5! to account for the 5 identical D's and by 4! to account for the 4 identical L's.
Thus:
Total possible routes = 9!/(5!4!) = 126.
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