I'm having a hard time grasping just how to apply my knowledge of counting methods to certain questions in my textbook. Although everything (seems to) make sense, the wording and layout of questions for this topic has a way of jumbling my understanding.

As I looked through the review section of my textbook only every second or third question I read I knew how to approach, despite them being on par in terms of difficulty with the others (as per the layout of the book). If you could help me with these questions, and maybe help patch the hole in my understanding along the way, that would be great.

1.Seven people are to be seated in a row. Calculate the number of ways in which this can be done so that two particular persons, A and B, always have exactly one of the others between them.

I know that the total number of ways the 7 people can be seated is 7! = 5040, but where do I go from here?

2.Determine the number of possible permutations (from digits 1 through to 9) when the digits 1 and 2 are together but not necessarily in that order.

First, I calculated the total number of permutations to be 9! = 362880. I figured that if 1 and 2 are together, we can count that as 1 digit; "12" or "21". Therefore, we now have only 8 digits in the set. The total way the new combined digit can be arranged is 2! = 2. I don't know where to go from here. I tried 9!/2(8!) but got 42 (the answer is 48).

3.There are 10 chairs in a row:

a) In how many ways can three people be seated?

10 x 9 x 8 = 720.

b) In how many of these will the two end chairs be occupied?

c) In how many of these will the two end chairs be empty?

4.All possible three-digit numbers are formed from the odd digits {1, 3, 5, 7, 9},

a) How many such numbers are possible if each digit is used only once?

5 x 4 x 3 = 60

b) How many of the numbers from part a are lager than 350?

I know that in order to achieve that, the number must start with either a 3, 5, 7, or 9, but I don't know how to go about that mathematically. I tried removing 1 from the set, leaving me only 4 numbers, and then doing 4 x 3 x 2 = 24, but was wrong.

Thanks for your help, it is greatly appreciated.