 Let's consider a set of n items. We want to create pairs of items. Assuming that we will not allow repetitions within the pairs, so we will not allow a pair to contain two items which are the same, and that we don't care about the order, so a pair AB is the same as a pair BA, then the number of pairs of items that we can create is n times n minus 1 divided by 2. Consider an example where we have four colored balls, red, green, blue, and yellow, so n equals 4. We have four items. Then the pairs that we can create with those four colored balls, red, green, red, blue, red, yellow, note that green, red is not a different pair, it's the same as red-green, so we don't count green-red as a separate pair. We can also create green-blue, green-yellow, and blue-yellow, and of course we cannot create a pair such as blue-blue, we do not allow repetitions. So in this case we have of the four items, four times three divided by two or six pairs. A second example, consider a communications network which has five stations, five computers, and we want to connect them together via links. So here we have five stations, A through to B, then to connect all possible pairs of stations we need to connect A to B, a link from A to C, A to D, A to E, so there's five links, and then B also needs links to the other station. So B already has a link to A, assuming that's a duplex link, B can send to A, so we do not count another link there, but B to C, B to D, and B to E. So sorry, the first set was four links. Now an additional three links gives us seven links, and of course C needs a link to D and to E, so an additional two links brings us up to nine links, and finally we need a link between E and D. So we have ten links in this case, and the equation in this case we have five items, five computers in this case, the number of pairs we can create is five times four divided by two, ten pairs, or ten links in this case. If we extended this network to a larger network which had ten stations, we will not draw it, but ten stations and the number of links we'd need would be ten times nine divided by two, which is 45 links. As a second example, let's say we have symmetric key encryption amongst a set of users, and in our system we have 50 users, so 50 people, and with symmetric key encryption each user may want to encrypt data to send to another user, so each pair of users needs a shared secret key. So how many keys do we need in this system with 50 users? The number of keys, well we need a key for each pair of users, so user A and B needs a key, so we can calculate the number of keys as 50 times 49 divided by two, which is 1,225. In a system of 50 users where each user uses symmetric key encryption, we need a total of 1,225 keys to be shared amongst those users.