How do you calculate this kind of...thing?

Question

When you have a certain number of digits, and these digits start and end somewhere, how do you know the maximum number of variations these digits can form?

For example, suppose I own a company and I was given a fixed IP address (the two parts to the left are fixed) and open (the two right parts are open). Suppose the left parts are 111.111 and the rest is open, for instance: my machine would be 111.111.001.001, my secretary would be 111.111.001.002, and so on. The two parts to the right start with 0 and end with 9. How can I know the maximum number of computers my company can have based on that very address?

Another way to phrase this question is this: suppose I have a safe and it has three digits that you have to get right for it to open. Suppose the permitted digits are 0-9 and A-F. It can be 001, 0F5, F79, and so on. How can I know how many variations this formula can form? I'm sure it would be a huge number, and that's what I want to know: how can I calculate that number?

Thanks

Answer 1

GO TO http://en.wikipedia.org/wiki/Combinations for the formulas>>>>

In combinatorial mathematics, a combination is an un-ordered collection of unique elements. (An ordered collection is called a permutation.) Given S, the set of all possible unique elements, a combination is a subset of the elements of S. The order of the elements in a combination is not important (two lists with the same elements in different orders are considered to be the same combination). Also, the elements cannot be repeated in a combination (every element appears uniquely once). This is because combinations are defined by the elements contained in them, so the set {1, 1, 1} is the same as {1}. For example, from a 52-card deck any 5 cards can form a valid combination (a hand). The order of the cards doesn't matter and there can be no repetition of cards.

Answer 2

I'm not sure of this answer, but it seems in this situation a combination would be appropriate. You have a choice of 14 different digits and only need 3. 14C3=364 combinations. Sorry if I am wrong...

Answer 3

For your first question, if 000.000 is permitted, you have 10^6 possible address. For your 2nd example, you have 16^3 possible combinations. In general, when you have m values which can be placed in n positions, and repeated values are allowed, N = m^n

Answer 4

according to my calculations the possible number of combinations is 5,225,472-if you assign the letter A a value of one-it's 1x2x3x4ect--A(1)xB(2) ect--then multiply both results--see if i'm right i'd like to know your answer--disregard zero, because anything multiplied by 0=0