 Alright, so let's go ahead and throw down some formulas, and again there are formulas for computing the number of permutations and combinations, but they are almost never useful. Now in order to have any chance of using the formula, we have to introduce the following idea. An important distinction is whether our choices are made with or without replacement. Making our choices with replacement allows the same choice to be made more than once. If we make our choices without replacement, this does not allow the same choice to be made more than once. So let's consider that, let's determine whether the selection is done with or without replacement. The digits of a phone number, the letters of a word, or the members of a student basketball team. So the digits of the phone number are chosen from the digits 0 through 9, and since a digit can be repeated, the selection is done with replacement. The letters of a word are chosen from the letters A through Z, and again since a letter can be repeated, the selection is done with replacement. The members of the team are chosen from among the students. Now in this case, since you can't choose the same person twice, the selection is done without replacement. So let's consider a generic permutation problem. Suppose we select n objects from m objects without replacement, how many permutations are there? So let's consider this as a sequence of choices. We make our first, second, third, and so on up to our nth choice. Now since we're selecting from a set of m objects without replacement, we have m choices for that first object. For our second choice, there's only m minus 1 objects left. For our third choice, m minus 2. And for the most complicated part of the problem, for the nth choice, we have m minus n plus 1 choices. Here you might notice that adding the choice number and the number of choices always gives you m plus 1. And our fundamental counting principle says that we can find the number of permutations as the product m times m minus 1 times m minus 2 all the way down to m minus n plus 1. Now let's see if we can find a formula for this. So remember that we define factorial n to be the product of the whole numbers from 1 through n. But here we have the product of the whole numbers from m minus n plus 1 up through m. So this is not a factorial. But wait, remember you can get anything you want to as long as you pay for it. If I want to have the product of the whole numbers from 1 up to m, let's go ahead and put that in. And in this case, notice that what we're really missing here is the product of 1 by 2 by 3 all the way up to m minus n. And we'll pay for it by dividing by that same set of factors. And if we look at this, our numerator is m factorial and our denominator is m minus n factorial. And so this leads us to the following result. The number of permutations of m objects where n are taken without replacement is given by the formula. Now there's a couple of variants on this formula. For example, this might be written this way. So let's check it out. How many three letter words can be formed by selecting letters of the alphabet? Since we can pick a letter more than once, we can't use our formula which only applies to permutations without replacement. Instead, we use the fundamental counting principle. We have our choice of first letter, second letter, and third letter. And we have 26 choices for that first letter, 26 for the second, and 26 for the third, or a whole lot of possibilities. But can't we use our formula? Well, sure. We can use our formula as long as we write a question that is designed to let us use the formula. And this is why these formulas are not all that useful because the only times we can use them are in problems that are specifically written to allow us to use the formula. So how many three letter words can be formed by selecting letters of the alphabet where no letter can be used more than once? Since we're not allowed to reuse letters, this selection is done without replacement and we can use our formula. Now we'll do this in both ways with the formula and without the formula. Now the old way, we had a choice for first, second, and third letter. There are 26 choices for the first, 25 for the second, and 24 for the third, so we get a lot of possibilities. But we have this nice new formula that applies to this situation. So let's use the new way. Since we're selecting three letters from 26 without replacement, we want to compute 26 permute 3. And our formula says that's 26 factorial divided by 26 minus 3 factorial or 26 factorial divided by 23 factorial. We'll calculate 26 factorial and 23 factorial and divide. Now actually computing those factorials can be a rather horrifying task, so let's see if we can do this a better way. And again, since we're selecting three letters from 26 without replacement, we want to compute 26 permute 3. Now before we actually calculate that, let's think about this. 26 factorial is the product of the numbers from 1 up to 26, or we might start at 26 and count down. Meanwhile, 23 factorial is a product of the numbers from 1 to 23, or from 23 down to 1. And the thing to notice here is that all of these factors from 23 down to 1 appear in both numerator and denominator. And so we can remove the common factors and we're left with. Now the thing to recognize here is this computation we get to is exactly what we've been doing when we did things the old way. So if we're going to be doing that, we might as well keep doing what we've been doing and not bother with this new formula.