Problem involving combinations and permutations

Hey all,
The following question was covered in a logic class, its somewhat fairly easy but I am finding it hard to grasp the concepts.

So let be an alphabet, I want to find how many elements they are in such that and occur exactly once.

Well there are 5x4=20 combinations of positions for the 2 letters to appear within a string of 5 characters, and then the other 3 characters have 2 possibilities each, and 2 to the 3rd power is 8 so it should be 160, I think. Probably not the way they want you to solve it though.

Thanks zortharg, but what if occurs exactly twice and occurs at least once..

Perhaps I should elaborate on why I said what I said. You get ONE single alpha. There are 5 possible choices as to where to put it. Then you get ONE single gamma. There are 4 places left you can put it. Thus the 5 x 4 = 20 possible choices. The remaining 3 spaces each have TWO choices, since the letters you were allotted one and only one occurrence of have already been used to capacity. Each of those 3 spots must have either a beta or a delta, but we don't care if beta gets used none, once, twice or thrice, delta getting used whenever beta is not. It is plainly 3 bits of information there. Each one can be either of the 2 with no other restrictions. Thus 8 possible choices for the remaining 3 spaces. That's how I got 20 times 8 or 160. So you see, the way I got my answer doesn't ALLOW for the possibility that alpha occurs twice or any number other than 1, and delta doesn't matter, who cares about delta.

I just realized you were asking a completely separate question. Trying to get someone else to do your homework for you, eh? I just completed part a and now you want the answer for part b, do you?

Well fine. Maybe at some point you'll get so embarrassed that you'll just do it yourself.

If alpha occurs twice, then that's 5 choose 2, or 10 ways you can pick 2 of the 5. For the remaining 3 characters, you have 3 letters you're allowed to use, one of which at least must be delta. Well, there are 27 combinations of 3 characters with 3 letters, but 8 combinations of 3 characters with 2 letters - 8 of the 27 are no good, in other words, since they don't make use of delta SOMEwhere. So that leaves 19, times the 10 ways you can select the 2 positions for alpha, for a total of 190 possible sequences.

Hey zortharg, Yea I missed the lecture that covered combinations and permutations. Got exam in 2 days, so going through tutorials at the moment, unfortunately they do not contain solutions.