Suppose in a competition 11 matches are to be played ,each having one of 3 distinct possible outcome .The number of ways one can predict the outcome of 11 matches such that exactly 6 turn out to be true is - (A) ${{11}\choose{6} }2^5$ (B) ${11\choose 6}$ (c) $3^6$ (D) None of the above.
I think this because $X=$ the random variable that count the number of correct prediction is binomial distributed . Hence probability such that exactly 6 out of the 11 prediction is correct= ${{11}\choose{6} }\frac{1}{3^6} \frac{2^5}{3^5}$ .
Hence Total number of favorable cases = required probability $\times $ total number of cases = ${{11}\choose{6} }\frac{1}{3^5} \frac{2^5}{3^5} \times 3^{11}= {{11}\choose {6} }2^5$ .
So answer is (A) .
Strictly speaking there is no probability/randomness mentioned in the problem, but you have the right idea, assuming predictions are made uniformly at random and independently for each match.
Another way to think about it is to just count the number of ways to correctly predict: there are ${11\choose 6}$ ways to choose which matches are correctly predicted. For each of those, there is no choice of outcome - it must be the correct outcome. For the other $11-6=5$ matches, you must predict incorrectly, so there are $2$ possible outcomes for each. Thus ${11\choose 6}2^5$.