We have $11$ students, one of them George. We find the probability that George gets exactly $5$ chocolates, on the assumption that the chocolates are given out to students, one at a time, independently and "at random." A success, at least from George's point of view, is that George gets the chocolate.
The probability of success on a given trial is $1/11$, the probability of failure is $10/11$. The probability of $5$ successes )and therefore $3$ failures) in $8$ trials is
That is precisely one of your solutions.
If the distribution of chocolates happens as described in the solution above, then the other solution is not correct. It is easiest to see that under that probability model, the answer is not correct. If we express the answer as a reduced fraction, it will have a high power of $11$ in the denominator. Your first answer, when expressed as a reduced fraction, does not have an $11$ in the denominator. (An $11$ comes from the top of the expression you gave, and one from the bottom; they cancel.)
As @joriki has written in his solution, the probability model was left unmentioned by the question setter. That is a serious error. For then, in principle, we can choose any model we like. Assume, for example, that the chocolate giver promised George that he would get $5$ chocolates. Then the required probability is $1$. Or else assume "randomness," except that someone must get $5$. then the probability is $1/11$. But presumably the question setter had some default model in mind. The most likely default assumption is the one described in my solution above. It is the same as the model used in your second solution.
Now we need to worry about why the other solution is not correct under our probability model. That is a little difficult, since I do not fully understand that solution. It looks like a variant of a "stars and bars" argument, though the particular numerical details are not explained. Stars and bars arguments are very good for counting the number of ways that $k$ identical objects can be distributed into $n$ bins. But that is often not useful for probability calculations, since the ways are in general not equally likely.
Imagine, in our stars and bars picture, that George will get all the chocolates up to the first separator. We can count the number $S$ of ways in which George can get $5$. We can also count the total number $N$ of ways to distribute the chocolates. The answer to our probability problem, under the default assumption, is not $S/N$, since the $N$ ways are not all equally likely. We are so accustomed to equally likely situations that it is easy to forget to check whether our sample space consists of equally likely outcomes.