How many ways to reach a given tennis-score? Let's say a tennis player wins a set with a game score of 6-3. In how many different ways can we reach this score?
Assuming H means the home-player won the game and A means the away-player won the game, one permutation would be HHHHHAAAH.
Note that the winner of the set will always have to win the last game (so maybe I could reduce this to how many permutations are there of the set of HHHHH AAA?)
I realize I have almost solved the problem but I'm still having problems grasping it, any ideas or if you could point me in the correct directions would be helpful.
For those unfamiliar with the tennis scoring system, a set is a sequence of games.  The set score is simply a count of how many games each player won.  For purposes of this question, all you need to know is that the first person to win a total of 6 games wins the set.
 A: As you saw yourself, you’re looking for the number of permutions of five H’s and three A’s. That’s a string of $8$ symbols, and the A’s can occupy any $3$ of the $8$ positions, so the desired number is simply the number of ways to choose $3$ things from a set of $8$ things. This is the binomial coefficient 
$$\binom83=\frac{8!}{3!5!}=\frac{8\cdot7\cdot6}{6}=56\;.$$
A: You're almost home! You've correctly reduced the problem to "how many ways can we introduce three As into a five Hs". So, of the eight positions which we can label $1$, $2$, $\dots$, $8$, three positions must be As, so a score stream such as HHAAHHAH might be represented as $\{3,4,7\}$. 
Hence, we can look at it as "how many $3$-element subsets are there of an $8$-element set". 
A: Those are called Permutations with repetition in general, repetitions are taken care of by dividing the permutation by the factorial of  the number of objects that are identical.So you have $5$ letters $H$ and $3$ letters $A$ hence the number of permutations is $$\frac{8!}{3!\cdot 5!}=56$$     
