4. Casting problem

- Earlier, you created a small crowd of robots by choosing just a few of the possible robot combinations. Let's back up a bit and go back to the example where robots consisted of one head, and one body. If you have two heads, and two bodies, then you can make two times two, or four, different robots. That's like having four possible actors to choose from but you only need a cast of three in your movie. Let's give our actors names: Alice, Bob, Carol, and Dave, or A, B, C, and D, for short. That leads to an interesting question: how many different casts of three actors can you make when you have four actors to choose from? Remember that, to figure out how many ways there are to combine our actors, we need to multiply our choices at each stage. In this case, we have four choices for our first actor, three for the second, and two for the third. So, it seems like we have 4 x 3 x 2, or 24, possible casts. Let's list out all 24 combinations. The first combination, ABC, means we selected Alice then Bob and then Carol, but there's a subtlety lurking in there. There aren't 24 different casts. To see where the subtlety is, notice that the second combination, ACB, means we selected Alice then Carol and then Bob. So, the first two combinations use the same actors, just in a different order. The same is true for the other combinations in the first box. It's the same cast. Just the order that we picked from was different. In other words, all the combinations in the first box should be counted as just one cast. Similarly, the second box is a cast consisting of Alice, Bob, and Dave. The total number of casts is, therefore, the number of boxes. So, how many boxes are there? Since there are six combinations in each box, there must be 24 divided by six, or four, boxes. So, there are four casts. But why is it that each group contains exactly six sequences? Why not three or four or any other number? The next interactive will help you visualize this problem using a few other examples. Have fun!