 If we are given a set of n items, and we're trying to create sequences of the length of k items, then there are n to the power of k possible sequences. This assumes that we're allowed to have repetitions of those items in each sequence. Let's consider some examples. First, let's say we have our set of items being the letters A, B and C. In this case, there is n equal to three items. We want to create sequences out of these three items of the length of two. So k is equal to two. That is, create a sequence containing two items, choosing from the set of A, B and C. We'll go through and list them all. We have A followed by A. That is, we're allowed to have repetitions of that item. We're going to have A followed by B, A, C. Well, the first item could be B, B, A, B, B, which includes a repetition, B, C, or C, A, C, B, C, C. There are no other possible sequences that we could create from those three items, and that gives us a total of nine possible sequences. And we see three to the power of two is nine. As another example, let's say that we, a user gets to choose a pin for a bank card, and they must choose a four-digit pin. So a four-digit pin, then there are ten possible digits to choose from for each of those digits we use in the pin. That is, N equals ten in this case. The digits are, of course, zero through to nine. And so in this case, when we choose the first digit, we can select from ten. So we can select zero through to nine. And then when we select the second digit of our pin, we can also select from zero through to nine. We're allowed to have repetitions and so on for the third and fourth digit in the pin. So in this case, K is equal to four, and the number of possible pins is ten to the power of four or ten thousand. And in fact, you can think about that. If we select our four-digit pin, we could choose zero, zero, zero, zero, or zero, zero, zero, one, zero, zero, zero, two, and so on. So the fourth digit in this case can be one of ten possible values. And then we could do the same, but where we vary the other digits. And that would go down to the last possible value of nine, nine, nine, nine, ten thousand possible values of our pin. As another example, let's consider a keyboard. A standard keyboard, an English keyboard, would normally have the set of letters, the set of twenty-six letters. And we want to consider, well, given the standard set of printable characters that we can use, how many can we choose from? In particular for choosing or typing in a password. So let's first look at the keyboard and consider that on the keyboard there are twenty-six letters. And of course, using the shift key, we can have uppercase or lowercase letters. So in fact, it gives fifty-two possible characters there. There are the ten digits. So that brings us up to sixty-two possible printable characters. But there's also some punctuation characters and operators. So the ten punctuation characters above the digits, if we use the shift key. Plus those other characters shown in green there. And that totals up to thirty-two punctuation characters and operators. Which typically means on a keyboard, for the characters that we can use to enter in a password, there are ninety-four characters to choose from. So given that, if we know that there are ninety-four printable characters to choose from on a standard keyboard, then consider a question, what if a user needs to select a password of length eight? So a password selection scheme forces the user to choose a password with eight of those characters. They can only choose from that set of ninety-four. How many possible passwords could be selected if they choose randomly or any possible combination? Well, we have N, the set of items N equal to ninety-four. And we choose sequences of length eight. So the number of possible passwords is ninety-four to the power of eight. And with a calculator, ninety-four to the power of eight is this number, which is approximately, if you look at it, six by ten to the power of fifteen. Or six thousand billion possible passwords.