Take the tour ×  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.
share|improve this question add comment  1 Answer  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
$\binom{11}6\cdot2^5$.  share|improve this answer add comment
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