1. Discrete Maths "Assortment" Set

Hi everybody,

First of all, I'm sorry, I'm afraid this will be a long post. =P And now a disclaimer: I'm posting here a set of three problems, but before people come bashing on me saying that I should do my homework myself, I'll let it clear from now that you can help in as many problems as you want.

OK, so for the issue itself. These problems come from my last test, and thing is, you can solve them using recurrence relations, generating functions or inclusion exclusion principle (you can also solve via traditional combinatory reasoning, worth less points, obviously). For didatic purposes, I'm asking hints for all three methods in every question (so, this will - hopefully - be a long topic too =P).

So for the problems themselves:

1) During 6 days of the week (from monday to saturday), a young man spends two afternoons in front of the computer, two in front of the TV and two in front of the video game. He spends the whole afternoon in front of only one equipment. Considering he doesn't spend two consecutive afternoons in front of the same equipment, in how many different ways can he program the 6 afternoons of his week (i.e., attributing equipments to the 6 aternoons)?

2) A young man drinks two soda bottles in each one of the afternoons from monday to saturday. Supposing he buys 6 identical bottles of soda brand A and 6 identical bottles of soda brand B, determine the number of ways to distribute the 12 bottles in the 6 days (two a day).

3) Each soldier of a group of $\displaystyle n$ must get a bubble gum. There are three available brands of gum, A, B and C, and the number of distributed gums from brand A must be even. Determine the number of ways to make such a distribution.

PS: my title doesn't make any sense, please replace "Assortment" with "Assorted".

2. 1) During 6 days of the week (from monday to saturday), a young man spends two afternoons in front of the computer, two in front of the TV and two in front of the video game. He spends the whole afternoon in front of only one equipment. Considering he doesn't spend two consecutive afternoons in front of the same equipment, in how many different ways can he program the 6 afternoons of his week (i.e., attributing equipments to the 6 aternoons)?
I suspect a more elegant solution is possible, but anyway

Call the equipment used on the first day A, the equipment used on the second day B and the equipment not used in the first 2 days C.
If A is used on the third day then the remaining 3 days are CBC, giving 1 possibility.
If C is used on the 3rd day then A or B can be used on the 4th day and there remain C and A or B (whichever was not used on the 4th) for the fifth day. This gives 2X2 = 4 possibilities.
Add this to the 1 from before and multiply by the number of permutations of equipment as A, B and C

2) A young man drinks two soda bottles in each one of the afternoons from monday to saturday. Supposing he buys 6 identical bottles of soda brand A and 6 identical bottles of soda brand B, determine the number of ways to distribute the 12 bottles in the 6 days (two a day).
partition the set by the number of days on which he drinks one of each soda.

3) Each soldier of a group of must get a bubble gum. There are three available brands of gum, A, B and C, and the number of distributed gums from brand A must be even. Determine the number of ways to make such a distribution.
Edit: Ignore this, it doesn't work because It only counts distributions where all As are next to another A

You can add a B or a C to the end each distribution of size n-1 or AA to the end of each distribution with size n-2, giving the recursion $\displaystyle t_n = t_{n-2}+2t_{n-1}$

3. 1) During 6 days of the week (from monday to saturday), a young man spends two afternoons in front of the computer, two in front of the TV and two in front of the video game. He spends the whole afternoon in front of only one equipment. Considering he doesn't spend two consecutive afternoons in front of the same equipment, in how many different ways can he program the 6 afternoons of his week (i.e., attributing equipments to the 6 aternoons)?
It seems to me that this question is like seating 6 people in a row table with 3 pairs who don't want to sit with their partner.

Now seating 6 in a row is 6!
Seating one pair together is 5! x 2!(count the pair as one, like seating 5 in a row with both arrangements of the pair)

There are 3 of them so 3 x 5! x 2

Seating two pairs together = 4! x 2! x 2! (count the pairs as 1)

There are 3C2 ways of choosing the 3 pairs so 3 x 4! x 2! x 2!
Seating three pairs together = 3! x 2! x 2! x 2!

One pair together less two pairs together less Three pairs together gives the number of ways pairs can be together.
Total arrangement of 6 less the above should give the number of ways not together.

6! - (5! x 2 - 4! x 4 - 3! x 8) = 624

The solution I though is to reduce the problem to counting the number of different ways to build a "word" with 6 As and 6 Bs (which would be $\displaystyle P(12;6,6) = \frac{12!}{6!6!}$, I guess). Is this correct?