 If we are given an n-bit binary number, then there are two to the power of n possible values of that number. As some simple examples, if we have a one-bit binary number, then the binary values are zero and one, or two values. If we have a two-bit value, then we have a two-bit number, then we have four possible values. If we have a three-bit binary number, then there are eight possible values, and so on. As an example, consider a protocol that has a sequence number, and the sequence number is incremented for each packet that it sends. The sequence numbers are usually fixed in size, so let's consider a sequence number which is stored as a 16-bit value. Then the number of possible values of the sequence number is two to the power of 16, which if we get our calculator out, two to the power of 16, 65,536. In protocols, sequence numbers are usually incremented, so if we start with a decimal value zero for the first packet, the second packet will be one, then two, and so on. And with 65,536 possible values, the largest possible value is 65,535. Since we started numbering at zero, the number of possible values is our two to the power of 16, with 16 bits. And with a sequence number, normally the next packet would wrap back around a zero, one, and so on. As a second example, consider IP addresses, in particular IP version 4. IP version 4 addresses are 32 bits long, although we often see them in dotted decimal form. The raw form is in binary, with 32 bits, there are two to the power of 32 possible IP address values. How many is that? We can calculate 2 to the power of 32 is in fact 2 to the power of 2 times 2 to the power of 30. And we can approximate that 2 to the power of 2 is in fact 4. 2 to the power of 30 is about 10 to the power of 9. So the approximate number of IP version 4 addresses, in theory, is about 4 billion. As a third example, consider we have a security key, and that key is chosen to be 128 bits long. So a user chooses 128 bit value, then an attacker which is trying to guess the key in the worst case would have to try all possible values before they got the right one. How many possible values they'd have to try? Well, 2 to the power of 128. What is that approximately equal? Maybe we'll use the calculator. A different calculator, BC, 2 to the power of 128 is this large number in decimal. And I've checked before, 2 to the power of 128. It's approximately, we'll see, something times 10 to the power of 38. It's approximately 3 by 10 to the power of 38. So an attacker trying to guess the key would, in the worst case, have to make 3 by 10 to the power of 38 possible attempts.