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Question Number 138960 by Dwaipayan Shikari last updated on 20/Apr/21
$${X}\:{and}\:{Y}\:{stand}\:{in}\:{a}\:{line},\:{at}\:{random}\:{with}\:\mathrm{10}\:{other}\:{peoples}. \\ $$$${What}\:{is}\:{the}\:{probability}\:{that}\:{there}\:{will}\:{be}\:{exactly}\:\mathrm{4}\:{persons}\: \\ $$$${between}\:{X}\:{and}\:{Y}? \\ $$
Commented by mr W last updated on 20/Apr/21
$${p}=\frac{\mathrm{2}×{C}_{\mathrm{4}} ^{\mathrm{10}} ×\mathrm{4}!×\mathrm{7}!}{\mathrm{12}!}=\frac{\mathrm{2}×\mathrm{25401600}}{\mathrm{12}!}=\frac{\mathrm{7}}{\mathrm{66}} \\ $$
Commented by mr W last updated on 20/Apr/21
$${yes}.\:{i}\:{corrected}. \\ $$
Commented by Dwaipayan Shikari last updated on 20/Apr/21
$${Hmm}\:{sir},\:{but}\:{answer}\:{given}\:{as}\:\frac{\mathrm{7}}{\mathrm{66}}\:! \\ $$
Commented by I want to learn more last updated on 20/Apr/21
$$\mathrm{Please}\:\mathrm{sirs}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{Q138976} \\ $$