Let’s say there is a part number ‘ABC-D-E-F-G.’ The ‘ABC’ section describes the basic part, with the 'D-E-F-G" section describing aspects of that part. Each of the ‘D-E-F-G’ aspects may be varied at the customer’s request, with five options (D, D1, D2, D3, D4, ect.,ect.,) for each parameter being available. ‘D-E-F-G’ aspects may be varied individually, in combination, or in total. Each variation logically results in a new part number, ‘ABC-D1-E-F-G2’, for instance. So, how many potential part numbers can be derived from the basic part and is this a problem in permutations or combinations? What equation would be used to solve this kind of problem?
Thanks for the answer, Desmostylus,but I don’t understand where the number “26” comes from. Maybe I mis-stated the number of options as “5” when it should have been “4.” I included the original condition “D” in my example “D, D1, etc., etc.” Also, the basic “ABC” portion of the part number would never change; only the “D-E-F-G” aspects are variable. Sorry to be thick-headed in case that’s what I’m being.
26 is the number of letters in the alphabet. If you have a code of the form “ABC”, there are 17,576 possibilites: AAA, AAB, AAC… ZZX, ZZY, ZZZ. Forget about that part, it was only there for illustration.
The next part is the “D-E-F-G”, where D, E, F and G can each take on any of 5 values. There are four “digits”: D, E, F and G.
You can make 5[sup]4[/sup] = 625 different enumerations.
Okay, thanks—I think I understand. There was a thread similar to this about a year ago, but I haven’t been able to find it. A similar question was answered in that thread, and I remember there was some discussion regarding the “permutation or combination” question. I don’t remember your approach being discussed, but your answer certainly seems logical. IIRC, treating it as a permutation resulted in something like 60K(?) possibilities, which seemed like much too large a number but I couldn’t refute the finding. At any rate, I do appreciate your input.
Desmostylus is correct, if the “ABC” part of the “number” stands for a string of any three letters. (And then, of course, there would be 17576 x 625 = 10,985,000 possible part numbers altogether.) When I first read the question, I half-assumed that you were using the letters ABC to stand for arbitrary numerical digits (0-9), which would make the number of possible basic parts 10 x 10 x 10 = 1000. And if each of the ABC could be a letter or a digit, there’d be 36 x 36 x 36 = 46656 possible parts.
And yes, this question doesn’t involve permutations or combinations, but what some textbooks call the “fundamental counting principle” or “multiplication rule for counting”: that when you have a series of choices to make, you get the total number of possibilities by multiplying together the number of ways to make each individual choice.
LouisB, I don’t think that we have fully understood your problem as yet. Please explain it more carefully. You say that there is a part named ABC. There are a number of variations on it. Each of these variations can be labeled ABC-D-E-F-G. How many possibilities are there for D? How many possibilities are there for E? How many possibilities for F? How many possibilities are there for G?
Let’s say that there are m possibilities for D and n possibilities for E and p possibilities for F and q possibilities for G. Then the number of possibilities for ABC-D-E-F-G is m times n times p times q. As has already been stated, this has nothing to do with permutations or combinations. It’s just multiplying the number of variations.
I might be able to simplify the technique used to solve this problem.
Problems of this nature are solved by figuring out how many values each position can take, and then multiplying the number of values for each position to arrive at the total number of values.
For instance, if a part-number is defined as AB and A can have 26 values (the number of letters in the alphabet) and B can have 10 values (the digits 0 through 9). Then the total number of part-numbers would be 26 X 10 = 260.
Wendell Wagner, as I said earlier, there are a maximum of five possible values for each of the “D-E-F-G” aspects of the part. These five include the original, factory specified value, with four additional options IF it is desired to modify the original. Of course, each and every modification would result in a new part number, since the part would no longer be identical to the original. However, the “ABC” portion is not subject to variation, so an example of a complete part number would be “ABC-D1-E3-F2-G4” as opposed to the original part number of “ABC-D-E-F-G.” I think that Desmostylus explained it pretty well. Thudlow Boink raises a good point, though, since in the real world, a lot of part numbers are alpha-numeric. For my purposes, though, given the explanation already received, I don’t think they will present a problem.