- Start here!
- 1. Two headed robots
- Counting two-headed robots
- 2. Snake bots
- Building snake bots
- 3. Calculating factorials
- Calculating factorials
- 4. Casting problem
- Counting casts 1
- 5. Does order matter?
- Counting casts 2
- 6. Binomial coefficient
Why do we divide the number of combinations by the number of permutations? If we have 4 actors (A, B, C, D) and we need to pick 3, we can arrange them in different ways. But, we don't want to count the same group twice just because they're standing in a different order. So, we divide by the number of ways the 3 actors can arrange themselves.
Want to join the conversation?
- Am I the only one who doesn't get the "bingo! that's it!" moment she gets at min0:50. I don't see at all the relationship between what she explained about permutations some seconds before and the seemingly arbitrary formula she's presenting afterwards.(8 votes)
- Me: Doesn't do the practice and saves it for tomorrow because i'm lazy
Her: gOoD wOrK
Me: i- i didn't even do anything-(8 votes)
- At0:58, she says, "...to keep order from mattering...". What does this mean and how does it apply to the casting problem? In terms of cast members, we do not care about their order; they are just an unordered set of actors. In terms of the number of members in the cast, that is just a number of slots -- there is no order. What part is the order mattering part?(7 votes)
- one question what does ! this mean when its front of the number!(3 votes)
- Good work! I left you with this question: Why are there exactly six combinations of each cast when you select three actors from a group of four? To get a feeling for what's up, let's look at the first two boxes. Notice that in the first cast, or box, we have all possible orderings of A, B, and C. Mathematicians call each of these orderings a permutation. Now, the number of permutations is represented by the number of rows in each box. The same is true in the second cast involving A, B, and D. How many orderings, or permutations, are there of three things? We saw earlier that there are three factorial, or six permutations. Bingo! That's it! To keep order for mattering, we need to take the total number of combinations where order does matter and divide that by all the possible permutations. So when choosing three actors from four actors total, we can write our calculation as four times three times two over three factorial, which equals four. Let's do another example and figure out how many casts we can form with three actors from a pool of six actors. In this case, there would be six times five times four. We then have to again divide by three factorial to keep the order from mattering, leaving six times five times four over three factorial, which equals 20. So there are 20 different casts in this case. Use the next exercise to get some practice with some other examples. And perhaps you'll recognize our good friend M-O in the cast. (robotic sounds)