1
$\begingroup$

The results of 6 soccer matches(win,loss or draw) are to be predicted. How many different forecasts can contain exactly 2 correct results?

So 2 correct results and 4 incorrect results. First I can select 2 matches out of 6 in $_6C_2$=15 ways. Now for each of these 15 forecasts containing 2 correct results, the remaining 4 matches can be selected in 4! ways. So 15*4! forecasts are possible.
What am I doing wrong?

$\endgroup$
  • 1
    $\begingroup$ Once you have selected the two correct results, you have also selected the four incorrect results. They are just the four remaining results. Wait a sec – maybe there's a distinction between the two ways of being wrong about a prediction (since there are three possible outcomes to each match). So it should be $15\times2^4$. $\endgroup$ – Gerry Myerson Feb 2 '17 at 9:56
  • $\begingroup$ There are also several ways you can be right about the correct results. To be explicit about the selection of hte wrong results: once you've picked the two matches you get right, the number of ways you can four to be wrong out of the remaining four is 1. $\endgroup$ – Iadams Feb 2 '17 at 10:01
  • $\begingroup$ @GerryMyerson please explain the last part to me, where you said " maybe there's a distinction between the two ways of being wrong about a prediction (since there are three possible outcomes to each match)" Thanks. $\endgroup$ – bandit_king28 Feb 2 '17 at 10:04
  • $\begingroup$ If Norway and Panama play to a draw, you can be wrong by predicting Norway will win, but you can also be wrong by predicting Norway will lose. So the four predictions you get wrong, you can get each of them wrong in two different ways. $\endgroup$ – Gerry Myerson Feb 2 '17 at 10:06
  • $\begingroup$ @GerryMyerson Got it. Thanks a lot! $\endgroup$ – bandit_king28 Feb 2 '17 at 10:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.