Solving probability and combinatorics questions methodically I have this problem:

Consider a deck of 52 cards (there are 13 cards of each suit).
a. How many different ways can you arrange, in a line, the 13 cards of
  the suit of diamonds, so that the cards 7,8,9 and 10 stay together
  either in the beginning or the end of the line?
b. 8 cards are extracted from the deck. What is the probability there is
  an ace and at least 3 queens in the extracted cards?

My questions:


*

*Is there any method to solve probability questions? 

*What strategies/lines of though do you use when tackling this kind of problems?

*Could you solve these problems using a method or several strategies and describe them?

*Is there any single method you can use to solve both of these seemingly different problems?
Thanks
 A: We will solve these problems, and see the techniques along the way.
Look at the first question. We are given $13$ diamond cards. The $7,8,9,10$ cards are supposed to stick together, either at the front or back of the line. Hence, we consider the possibilities separately:
1) The cards are at the front of the line. Then, these can be permuted amongst themselves in $4!$ ways, and the rest can be permuted amongst themselves in $9!$ ways so the the answer is $4!9!$
2) The cards are at the back of the line. Here, it is important to see a symmetry, namely that if you take any arrangement of the first case and reverse it, you get an arrangement of the second case, and vice-versa. Hence, the answer here is $4!9!$ again.
So the answer would be $2 \times 4!9!$
We come to the second question.
First of all, the number of all possibilities when $8$ cards are picked is $\binom{52}{8}$,without observing the order of picking. 
Next, we look at our favourable cases. We split them in two:
1) An ace and exactly three queens. The ace can be picked in four ways, and the three queens can be picked by excluding one queen, which is done in four ways again. We don't care what the rest are, these are picked in $\binom{48}{4}$ ways.
2) An ace and all four queens. The ace can be picked in four ways, and the four queens are fixed. We don't care what the rest are, these are picked in $\binom{48}{3}$ ways, because we have three remaining cards to pick out of $8$.
Thus, the answer would be $\dfrac{16 \times \binom{48}{4} + 4 \times \binom{48}{3}}{\binom{52}{8}}$.
The strategies used in these questions is very simple:
1) Split the favourable case into exclusive events i.e. events that cannot occur simultaneously. For example, $7,8,9,10$ cannot have been both the start and the end of the 13-card line, so we could examine them separately.
2) Know when we are permuting and when we are combining. The first question had to do with permutation, because in a line, the arrangement matters. However, In the second question, we do not care what order the aces,queens etc are picked in, so we have to take combinations.
3)The concept of symmetry is useful. It is nice to see when one situation can be transformed to the other symmetrically, as  I did with the reversing in the first question. I should add that the second case could have been dealt with separately, but the answer would be the same.
4)Exclusive events are added, independent events are multiplied. I hope you get the difference.
That's it. Follow these rules, you should be ok. 
A: (a) I'm going to assume the 7,8,9,10 have to remain in order here. If not, you have to multiply by 4! , which is the number of ways to have that 'straight' of 4 cards.
(1) First deal with the 13 diamonds. Place the 7,8,9,10 together in the front of the row and the rest behind them. How many ways can you place the rest of the diamonds behind the 7,8,9,10? There are 9 diamond cards remaining, you can permute them any way you like which is 9! ways (all those permutations are unique).
(2) Now place the 7,8,9,10 together at the end of the row. You can do this the same number of ways, the only difference being the location of your 'straight'.
2 * 9! is the number of ways to do this.
You can think of this in a different way if you'd prefer ... 
How many ways can you rearrange 9 cards? 9! ways. How many places can you place your straight? 2. So how many combinations of cards did you just consider? 2 * 9!
A: These are basic Combinatorics problems; those of the study of counting.  
The first just deals with permutations (aka arrangements).  You just need to count the ways to place the group of $7,8,9,10$, arrange it the group, and arrange the remaining nine cards in the suit.
The second you need to count the total space: selection of 8 cards from 52, and the favoured cases: select one from four aces, three-or-four from four queens, and four-or-three(respectively) from the forty-four other cards.
