Predicting the outcomes of a subset of chess games correctly 
Suppose $n$ games of chess are played. In how many ways can I predict the outcomes of $m$ of the games ($A$ wins, $B$ wins, there is a draw) correctly?

Here's my solution.
I can choose the $m$ games in one of $n \choose m$ ways. For each of these choices, there are $n - m$ games remaining whose outcomes I predict incorrectly, i.e. in one of the two outcomes that the game did not result in. There are two choices for each of the $n -m$ games, giving $2^{n-m}$.
Hence the answer is 
$$2^{n-m}{n\choose m}$$
Please tell me if there any mistakes here.
[Source:  Challenge and Thrill of Pre-college Mathematics, chapter Permutations and Combinations, exercise 9.2, problem no. 7]
 A: Oh, ok, the other problem @Soham that you link in a comment is more clear. This is a binomial distribution.
Think like this: you have a sequence of independent events. Any of them can take only two values: good or bad (if they are more than two possible values then is a multinomial distribution).
Then based in it "atomic" probability for a isolated event the probability for some sequence where a concrete number $m$ will be good and other $n-m$ bad will be:


*

*the probability for the event to be good (that is $p=\frac{1}{3}$, in this case to predict correctly the result of a game, powered to the number with correct predictions,i.e.,$(\frac13)^m$

*multiplied by the cases where the predict ISNT correct, what is $(1-p)=\frac23$, powered to it number of cases, i.e., $(\frac{2}{3})^{n-m}$

*this multiplied by the possible number of permutations of these two groups on the sequence, i.e., $\binom{n}{m}$


$$P(X=m)=\binom{n}{m}\left(\frac13\right)^m\left(\frac{2}{3}\right)^{n-m}$$
But the problem isnt asking for any probability for a concrete number of positive predictions... it is asking for the number of ways that you can predict it. The number of ways for $m$ correct predictions and $n-m$ incorrect is just the number of permutations,i.e.,$\binom{n}{m}$.
But, still, the question isnt enough clear... because I dont know exactly what is asking for. It can be interpreted, too, as the different values that can have the complete sequences (values: win, lose, draw). From this interpretation the number of possible predictions will be
$$\binom{n}{m}1^m2^{n-m}$$
But it can be interpreted too as the number of sequences that can lead to a prediction of $m$ correct predictions and $n-m$ incorrect. In this case is just the total number of possible different sequences, i.e. $3^n$.
