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I have found this question:

Suppose in a competition $11$ matches are to be played, each having one of $3$ distinct outcomes as possibilities. What is the number of ways one can predict the outcomes of all $11$ matches such that exactly $6$ of the predictions turn out to be correct.

My approach is: Any $6$ games(that will be predicted correctly) can be selected from $11$ games in $\binom{11}{6}$ ways and since each game has three possible results, there can be $\binom{11}{6}\cdot 3^5$ ways.

Please correct me if I am wrong.

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Your basic approach is okay, but you overlooked the fact that the other $5$ predictions have to be wrong. There are $\binom{11}6$ ways to choose which of the $6$ predictions are correct, but then there are only $2^5$ ways to predict the other $5$ outcomes incorrectly, not $3^5$, so the correct result is $\binom{11}6\cdot2^5$.

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