It was once again another week where I felt like I made a lot of progress and learned a lot but the amount of material I got through didn’t reflect it. In fact, I only got through one unit, Counting, Permutations, and Combinations which I was already ~30% of the way through when I started the week. I’m disappointed that I only got through the one unit since it wasn’t that big, but I do feel like I got a strong grasp on how permutations and combinations work, why they’re useful, and how to derive and interpret their formulas. Learning the factorial operation (the operation that’s used in both permutations and combinations) is like having learned about exponents or radicals; I can see why it may come in handy as I continue working towards understanding calculus.
I had a small epiphany at the beginning of the week that didn’t have much to do with either permutations or combinations. I can’t remember exactly how or why it happened, but I finally realized how to interpret a fraction (ex. A/B) with another denominator below it (ex. A/B/C). Here’s a page from my notes that breaks it down:
For lack of a better term, if you considered a regular fraction to have two layers (i.e. the numerator and the denominator), I never understood how to interpret a fraction with a third layer below it. I realized that in the example A/B/C both B and C are considered denominators and that you would divide the numerator by both of them. To simplify things, you can also multiply both denominators and divide the numerator by their product. In retrospect this seems completely obvious and something I should have figured out a long time ago but, regardless, I’m glad to have finally put it together. (At the bottom of the picture where it says “NOT: 24/3/2 = 24/1.5 = 16”, this is what I was confused about before now.)
Looking back at my post from last week, I don’t think I was correct in my terminology when talking about permutations and combinations. As well, in order to know the difference between permutations and combinations, it’s also necessary to know what the factorial operation is and how it works. Here’s what I have come to learn about all three things so far:
- Factorial
- An operation which takes any number and multiplies it by every integer less than it down to and including 1.
- Denoted with an exclamation mark, “!”.
- Ex. 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720
- Formula: n * (n – 1) * (n – 2) * … * 1
- Permutations
- The number of ways a set of items can be placed into a set number of ‘buckets’ when ORDER MATTERS.
- I.e. {A, B, C} ≠ {C, B, A}
- {A, B, C} and {C, B, A} are two separate ‘items’. (I’m also not sure if using those brackets to define each item is the right way to do it.)
- I.e. {A, B, C} ≠ {C, B, A}
- To figure out how many permutations of 6 items there can be in 4 ‘buckets’:
- Bucket 1 = (6) possibilities, bucket 2 = (5) possibilities, bucket 3 = (4) possibilities, and bucket 4 = (3) possibilities.
- Therefore, the number of total possibilities when order matters is 6 * 5 * 4 * 3 = 360.
- It can be written as 6!/(6 – 4)! which equals 6!/2! which is the same thing as saying 6 * 5 * 4 * 3 *
2 * 1/2 * 1.
- On a calculator, the function reads as “_nP_k” where n and k are subscripts.
- n = Number of items
- P = Permutation Function
- k = Number of ‘buckets’/’spaces’ the items are being sorted into.
- The number of ways a set of items can be placed into a set number of ‘buckets’ when ORDER MATTERS.
- Combinations
- The number of ways a set of items can be placed into a number of ‘buckets’ when ORDER DOESN’T MATTER.
- I.e. {A, B, C} = {C, B, A}
- {A, B, C} and {C, B, A} are considered the same ‘item’ and, for that matter, so would {A, C, B}, {B, A, C}, {B, C, A}, and {C, A, B}. All 6 orders are considered 1 combination.
- I.e. {A, B, C} = {C, B, A}
- To figure out how many combinations are possible when placing a number of items into a set number of buckets you figure out how many permutations there are and then divide by k!, i.e. the number of ways the ‘buckets’ can be rearranged since order doesn’t matter.
- Ex. the total combinations of placing 6 items into 4 buckets would be:
- 6!/(6 – 4)!/4!
- The 4! on the end of the equation is there because you must divide the numerator by the number of ways the buckets can be rearranged since the order of the buckets doesn’t matter.
- Ex. the total combinations of placing 6 items into 4 buckets would be:
- On a calculator, the function reads as “_nC_k” where n and k are subscripts.
- n = Number of items
- C = Combination (or maybe “choose”? I think I heard Sal say that.)
- k = Number of ‘buckets’/’spaces’ the items are being sorted into.
- The number of ways a set of items can be placed into a number of ‘buckets’ when ORDER DOESN’T MATTER.
The unit test was fairly straightforward and was only made up of two types of questions. One type of question gave me 3-5 groups of things, with 2-5 items per group, and asked me to find how many possible combinations were possible. An example of this type of question would be:
- Q. If John had 3 shirts, 4 pairs of pants, and 12 pairs of socks, how many different outfits could he create?
- Total outfits = 3 * 4 * 12
- = 144
- Total outfits = 3 * 4 * 12
As you might imagine, I found these questions very easy. The second type of questions were much harder. They were probability questions that required me to use permutations and combinations. The hardest part about these types of questions was figuring out what the question was asking and how to come up with the correct equation. I realized writing everything out as clearly as possible made things easier. When writing things out, I found it very helpful to write out the probability formula, Probability = (total # of ‘winning’ outcomes) / (total possible outcomes), and then fill in the blanks. Here’s an example of one of these types of questions:
- Q. If there were 8 students in a group, 2 boys and 6 girls, and the teacher picked 5 students at random, what’s the probability the teacher would pick all girls?
- P(5 girls) = (total combinations of 5 girls) / (total possible combinations)
- Total combinations of 5 girls = 6!/(6 – 5)!/5!
- = 6
- Total possible outcomes = 8!/(8 – 5)!/5!
- = 56
- P(5 girls) = 6/56
- = 3/26
- = ~0.1154
- = ~11.54%
- Total combinations of 5 girls = 6!/(6 – 5)!/5!
- P(5 girls) = (total combinations of 5 girls) / (total possible combinations)
As I said, the hard part was determining what values/equations needed to be placed in both the numerator and the denominator. After going through a few of these questions and getting used to writing everything out, it started becoming much easier.
There’s no chance I’ll be able to get through this course by the end of Week 52 but that’s ok. In fact, considering how big it is, I’m making my goal for the next two weeks to simply get through the single upcoming unit, Random Variable (0/2100 M.P.). I still think there’s a good chance I’ll be able to get through all of the statistics courses by the end of October. Oddly enough, each course seems to contain most of the same units. Although I won’t finish this course by the end of Week 52, I’m proud that I’ve stuck with this for an entire year and impressed with how much I’ve been able to learn. Two more weeks to go!