Now we're going to talk about how probabilities can be thought of as the probability of certain events occurring in a certain order, or the probability of a certain group of events occurring. So, for example, I might have five people working for me and I have five different assignments. And I need to place those people in those five different assignments. I can place the first person in any one of the assignments. So I have five people and I have five assignments. I can place the first person in five places. Now one square's occupied, so I can place the second person in four places, three spaces, two places, and now there's only one possible place to place the last person, okay? So, you see that there are 120 possible ways that I could put 5 people in 5 different jobs. Here the order of the assignment matters, so this is known as a permutation, okay? But let's say that I had five people and I wanted to form a team of five people who are all equal. There's no ordering, there's no specific role assigned to each person. Well, now there's really only one combination possible, I just need to assign my five people to the team. When order doesn't matter, there are always fewer possibilities, because multiple orderings of the same committee are accounted as one combination. Here we have 6 orderings of 3 distinct objects, so that would 6, 1, 2, 3, 4, 5, 6, permutations. But, the members of the group are constant, and so there's only one combination. Let's imagine a lottery starting with a bowl that has six different balls in it, each labeled. So they're labeled 1, 2, 3, 4, 5 and 6. And we're going to draw four balls, one at a time, to identify the winning number for the lottery. There are two ways you can think about the rules to this lottery. Either the order of the numbers matters or it does not. If the order matters, then we're talking about the number of permutations. The way four positions, one through four, can be filled by drawing from six unique objects. So, the first ball could be any of 1, 2, 3, 4, 5, 6, and having drawn those 6 there are 5 left, so we have any of 5 to go in the next place. And that leaves 4 to go in the next place, and that leaves 3 that could go in the last place. So, our total number of permutations would be 6 x 5 x 4 x 3 = 360. If you need to predict the order to win the lottery, then a fair bet would pay 360 to 1, or $360 on a $1 wager. The number of permutations has a general formula that can be written n factorial / (n-m) factorial. Where the exclamation mark stands for factorial and n is the number of unique objects, so n=6. And m is the number of unique attributes, or here the orderings from first to fourth. So m = 4, n- m would be equal to 2. So we would have 6 factorial / 2 factorial, which of course is equal to 6 x 5 x 4 x 3, or 360. On the other hand, if in order to win the lottery it doesn't matter what order we select the balls, we simply have to select the four correct numbers. Now we're talking about combinations. So let's imagine that I reach in here and I draw out the 4, the 2, the 3 and the 5, okay, 4, 2, 3, and 5. Now let's suppose that I only need to guess the correct four numbers, but not their ordering. In this case, I am going to divide 360 by 24 to get 15 possible combinations of 4 balls. And now, to win a lottery based on combinations, I would only receive $15 on a $1 wager if the lottery were fair. Because now I have a 1 in 15 chance of guessing the 4 numbers correctly. Why 15? Because there are 24 different ways that 4 balls can be ordered. So we need to divide 360 by 24. And (360 / 24) = 15. Why 24? Well I've written them out here for you. Because the first number, Second number, third number and the fourth number can be first. Then we have three numbers that could be in second place. And then for each of those combinations we have two remaining. So we have 4 x 3 x 2 or 24 possibilities. The formula that gives the general answer 15 can be expressed as n factorial / (n- m) factorial times m factorial. And this formula has a special name, it's called n choose m. And here we are showing the value n choose m for n = 6 and m = 4, it's 360. Divided by 24, which is equal to 15. Now let's consider another example. I need to send a dump truck, a bulldozer, and a steamroller to a construction project. And I have eight potential drivers, each of whom is qualified to drive all three machines. If I care about exactly which driver operates which machine, then I could have any of 8 drivers drive the first vehicle, 7 drive the second vehicle, and 6 drive the third vehicle. So, I have 8 x 7 x 6 = 336 unique ways that I could send out teams to drive 3 machines. So, here, the general formula is for permutations, which is n factorial divided by n minus m factorial, n = 8 and m = 3. So we have 8 factorial /(8-3) factorial, which is 5 factorial, and that would give us 8 x 7 x 6, okay? On the other hand, if it doesn't matter which driver drives which machine, I just want to know how many distinct teams of three drivers I might send to the construction site with my vehicles. In that case I use the formula for the number of combinations, which would be n factorial divided by n minus m factorial times m factorial. Or 8 choose 3, which would be 8 factorial / 5 factorial times 3 factorial. So I'm going to divide my previous answer by 6. And now I'm going to get 56 possible teams that I could send. And that is the difference between combinations and permutations. Next, we need to consider the concept of with replacement and without replacement when we're defining the probability of a certain situation. When we draw a card from a deck of cards, say in a poker hand, that card is removed from the deck and there's no longer any possibility of getting that card again until we have a new hand and a new situation. So for example, if I draw an ace on my first card, I would have a 1 in 13 chance of doing that. And my probability of drawing an ace on the second card would be different. There are now 3 aces left in the deck out of 51 cards. And so my probability of drawing an ace on the second card would be 3/51, okay? So any type of situation where we're removing things from the realm of possibility is without replacement, okay? On the other hand, if I'm trying to predict or guess a number between 000 and 999, it's certainly allowed to use each of the digits, 0 through 9, more than once. You can use each one two or three times, and so there we're generating a number and the digits are being used with replacement. So between the concept of permutation and combination and the concept of with replacement and without replacement, we have almost all probability situations that are likely to arise in a basic probability course.
