Eight videos, three minutes long each; enough time for five. How many combinations?

Question

Justin has 8 videos, all approximately 3 minutes long, but only has time to watch 5 of them. How many different combinations of these five videos can he watch?

$$5V3= \frac{5!}{(5-3)!}=\frac{5!}{2!}=\frac{120}2=60$$

Answer 1

Notice that Justin is selecting five of the eight videos, not three of five.

Let's distinguish between two questions.

In how many ways can Justin watch five of the eight videos?

Justin can choose the first video in eight ways, the second video in seven ways (from the remaining videos), the third video in six ways, the fourth video in five ways, and the fifth video in four ways. Hence, he can watch five of the eight videos in $$8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 = 6720$$ ways.

This is an example of a permutation problem since the order in which the videos are watched matters.

Observe that we could write the answer in the following form $$8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = \frac{8!}{3!} = \frac{8!}{(8 - 5)!}$$

In general, the number of ways of selecting $k$ items from $n$ items when the order of selection matters is $$P(n, k) = \frac{n!}{(n - k)!}$$

In how many ways can Justin select five of the eight videos to watch?

In this case, we do not care about the order in which the videos are selected, so we divide the answer above by the $5!$ orders in which Justin could select the same five videos. Thus, the number of ways Justin could select five of the eight videos to watch is $$\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 8 \cdot 7 = 56$$

Notice that we could express the answer in the form $$\frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} = \frac{8!}{5!3!}$$

This is an example of a combination problem since the order in which the videos are selected does not matter, just which subset of videos is selected.

The number of ways of selecting a subset of $k$ elements from a set with $n$ elements is $$C(n, k) = \binom{n}{k} = \frac{n!}{k!(n - k)!}$$