Counting

For all 3 problems below, Julia has exactly 6 hours to binge-watch episodes. You must show work that supports your final answer.

Part (a)

Assuming that for a given show she always watches episodes in sequential order (never out of order), how many orderings exist for her to watch 8 half-hour episodes from show A and 2 one-hour episodes from show B (she can switch between shows freely)?

Setup: The 8 half-hour episodes (A1, A2, …, A8) and 2 one-hour episodes (B1, B2) fill exactly $8 \times 0.5 + 2 \times 1 = 6$ hours. Since within each show the order is fixed, the only choice is which 2 of the 10 total episode slots are occupied by show B’s episodes.

Part (b)

Now assume she is willing to watch episodes of the same show in any order (not sequentially). How many orderings exist of the ten total episodes from part (a)?

Setup: All 10 episodes are now distinguishable and can appear in any order, giving:

Part (c)

She again watches in sequential order within each show. There are two shows C and D: one has at least 12 half-hour episodes, the other has at least 3 two-hour episodes. She can mix episodes from both shows in any order. How many orderings exist of the episodes she will watch?

Setup: Let $x$ = number of half-hour episodes watched and $y$ = number of two-hour episodes watched. Since she has exactly 6 hours:

Since episodes within each show are in order, the only freedom is choosing which $y$ of the $x+y$ total slots are two-hour episodes. Break into cases: