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$11$ matches are to be played,Each having $3$ distinct outcome, in how many ways one can predict the outcomes such that $6$ outcomes turn out to be correct?

My thought

$11C_{6}\times 3^5$

am I right?

Myshkin
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1 Answers1

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Hint There are $\binom{11}{6}$ different combinations of the $6$ different outcomes which can be correct.

Now, you only have $1$ for each of the $6$ matches, and you have $2$ different choices for the $5$ matches which need to be wrong (not 3).

Thus, there are exactly $\binom{11}{6}2^5$ different ways of picking exactly 6 outcomes.

If you need at least $6$ outcomes, calculate exactly 6,7,8,9,10 and exactly 11, otherwise you double count some combinations.

N. S.
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  • Each having $3$ distinct outcome was in the question, I think which is normal win/loose/tie – Myshkin Jan 12 '14 at 03:17
  • @Tojamaru And if there are 3 possible outcomes, how many are wrong? – N. S. Jan 12 '14 at 03:29
  • @Tojamaru Also, does the question asks for "exactly 6 outcomes to be correct" or "at least 6outcomes to be correct"? – N. S. Jan 12 '14 at 03:30
  • Exactly $6$ not atleast, and I do not understand your first comment – Myshkin Jan 12 '14 at 05:08
  • @Tojamaru You pick the $6$ right results in $\binom{11}{6}$ ways. The other $5$ have to be WRONG, so you don't have $3$ choices for each. The right result in any of the remaining $5$ games would give you a seventh right result. – N. S. Jan 12 '14 at 05:16
  • Do we have have to count separately that out of the $6$ choices, the first one is 'win' and that is correct and there are other two possibilities so we have to consider them as well? So in total $3 \times^{11}C_6 2^5$? – Ri-Li Jul 18 '19 at 17:24