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Question Number 103357 by bobhans last updated on 14/Jul/20
$${There}\:{are}\:\mathrm{21}\:{students}\:{will}\:{be}\:{trained} \\ $$$${with}\:\mathrm{6}\:{trainers}\:{available}.\:{Every}\:{students} \\ $$$${is}\:{trained}\:{by}\:\mathrm{1}\:{coach}\:{and}\:{every}\:{coach} \\ $$$${trains}\:{students}\:{with}\:{different}\:{amounts}. \\ $$$${many}\:{ways}\:{of}\:{grouping}\:{students}\:{who} \\ $$$${will}\:{be}\:{trained}\:{are}\:.... \\ $$$$\left({a}\right)\:\frac{\mathrm{21}!}{\mathrm{6}!}\:\:\:\:\left({b}\right)\:\mathrm{6}!.\mathrm{21}!\:\:\:\:\:\left({c}\right)\:\frac{\mathrm{21}!}{\mathrm{2}!.\mathrm{3}!.\mathrm{4}!.\mathrm{5}!\:\:}\:\: \\ $$$$\left({d}\right)\:\mathrm{6}×\mathrm{21}!\:\:\:\:\left({e}\right)\:\frac{\mathrm{21}!}{\mathrm{15}!}×\mathrm{6}! \\ $$
Answered by bemath last updated on 14/Jul/20
$$\Leftrightarrow\:\frac{\mathrm{21}!}{\mathrm{1}!.\mathrm{20}!}\:×\frac{\mathrm{20}!}{\mathrm{2}!.\mathrm{18}!}×\frac{\mathrm{18}!}{\mathrm{3}!.\mathrm{15}!}×\frac{\mathrm{15}!}{\mathrm{4}!.\mathrm{11}!} \\ $$$$×\:\frac{\mathrm{11}!}{\mathrm{5}!.\mathrm{6}!}×\frac{\mathrm{6}!}{\mathrm{6}!.\mathrm{0}!}\:×\mathrm{6}!\:=\:\frac{\mathrm{21}!}{\mathrm{2}!.\mathrm{3}!.\mathrm{4}!.\mathrm{5}!} \\ $$
Commented by bobhans last updated on 15/Jul/20
$${nice}\:!!\:\smile\therefore\smile \\ $$