Question Number 160389 by mr W last updated on 28/Nov/21
Commented by mr W last updated on 29/Nov/21
$${thanks}\:{for}\:{trying}! \\ $$$${but}\:{it}'{s}\:{wrong}\:{sir}.\:{i}\:{think}\:{you}\: \\ $$$${misunderstood}\:{the}\:{question}.\:{it}\:{is}\:{not} \\ $$$${to}\:{select}\:{one}\:{pair}\:{from}\:\mathrm{20}\:{people},\:{but} \\ $$$${to}\:{build}\:\mathrm{10}\:{pairs}\:{from}\:\mathrm{20}\:{people}. \\ $$$${the}\:{two}\:{students}\:{from}\:{any}\:{college}\: \\ $$$${should}\:{play}\:{against}\:{the}\:{students} \\ $$$${from}\:{other}\:{colleges},\:{not}\:{against}\:{each}\: \\ $$$${other},\:{so}\:{they}\:{should}\:{not}\:{be}\:{in}\:{a}\:{pair}. \\ $$
Commented by cortano last updated on 29/Nov/21
$$\mathrm{n}\left(\mathrm{S}\right)=\mathrm{C}_{\mathrm{2}} ^{\mathrm{20}} =\:\mathrm{190} \\ $$$$\mathrm{n}\left(\mathrm{A}\right)=\mathrm{190}−\mathrm{10}=\mathrm{180} \\ $$$$\mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{180}}{\mathrm{190}}=\frac{\mathrm{18}}{\mathrm{19}} \\ $$