Lena's Package Delivery

There are two cases: either Lena goes and returns on the same day or on two different days.

Case 1: If she goes and comes back on the same day, there are 3 + 2 + 1 (or 6) cases. There are seven days in a week; we now have 42 possibilities.

Case 2: If she goes and comes back on different days, there are no limitations with the 4 hour journey. She has 5 buses she can take going and 5 on any other given day to come back on. Let's say she goes on Sunday. We have 5 x 5 x 6 possibilities because she has five trains she could go on on a given day, 5 to come back, and 6 possible days she could do so. We see a repeating pattern... adding up 5 x 5 x 6 with 5 x 5 x 5 , 5 x 5 x 4 and so on, we get a total of 525.

Finally, 525 + 42 equals 567.

Solution 1: There are two cases: either Lena goes and returns on the same day or on two different days.

Case 1: If she goes and comes back on the same day, there are 3 + 2 + 1 (or 6) cases. There are seven days in a week; we now have 42 possibilities.

Case 2: If she goes and comes back on different days, there are no limitations with the 4 hour journey. She has 5 buses she can take going and 5 on any other given day to come back on. There are $7 \choose 2$ pairs of days that she can do this journey, hence there are a total of $5 \times 5 \times 21 = 525$ ways.

Finally, 525 + 42 equals 567.

Solution 2: If Lena takes a 6:00 am bus from Vancouver, there are 3 buses she could take home that day. If she takes a 9:00 am bus, there are 2 buses she could take that day. If she takes the 12:00 pm bus, there is 1 bus she could take home that day. If she takes the 3:00 pm or 6:00 pm bus she cannot return that day. So for each day from Sunday to Saturday (7 days total), there are 6 trips she could take that go and return on the same day. So in total there are 42 trips that she can take where she goes and comes back on the same day. If she returns on a day different from the day she left, she is able to take any of the buses those days. There are $\binom{7}{2} = 21$ ways to choose two different days during the week. Because Lena is in Vancouver, she must always go to Seattle on the earlier of the two days, and return on the later day. For each pair of days, there are 5 possible buses in one direction and 5 in the other, for a total of 25 possible trips for those two days. Over all pairs of days, this gives $21 \times 25 = 525$ possible trips. Combined with the 42 trips that return on the same day, there are 567 total trips Lena could take.

Solution 3: If Lena takes the 6:00 am bus to Seattle on Sunday, there are 33 buses she could take home. If she takes the 9:00 am there are 32. If she takes the 12:00 pm there are 31. If she takes the 3:00 pm or 6:00 pm bus there are 30. So if she leaves on Sunday there will be 156 different trips she can take. If she moves her departure ahead 24 hours, there will be 5 fewer buses she can take home (for each given starting time), so if she leaves on Monday, there will be $156 - 25 = 131$ trips Lena can take. Over the seven days, this gives us a seven-term arithmetic series with first term 156 and difference of -25, so the sum is $7 \times \frac{156 + (156 - 6 \times 25)}{2} = 567$ .