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Question Number 221661 by mr W last updated on 09/Jun/25
Commented by mr W last updated on 09/Jun/25
$${an}\:{unsolved}\:{old}\:{question} \\ $$
Answered by MrGaster last updated on 10/Jun/25
$$\mid{S}\mid=\mathrm{44} \\ $$$${k}=\frac{\mathrm{44}}{\mathrm{1}}=\mathrm{11} \\ $$$$\begin{pmatrix}{\mathrm{44}}\\{\mathrm{4}}\end{pmatrix}\begin{pmatrix}{\mathrm{40}}\\{\mathrm{4}}\end{pmatrix}\begin{pmatrix}{\mathrm{36}}\\{\mathrm{4}}\end{pmatrix}\ldots\begin{pmatrix}{\mathrm{1}}\\{\mathrm{4}}\end{pmatrix}=\underset{{i}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}\begin{pmatrix}{\mathrm{44}−\mathrm{4}{i}}\\{\mathrm{4}}\end{pmatrix}=\frac{\mathrm{44}!}{\mathrm{4}!\mathrm{40}!}\centerdot\frac{\mathrm{40}!}{\mathrm{4}!\mathrm{36}!}\centerdot\frac{\mathrm{36}!}{\mathrm{4}!\mathrm{32}!}\ldots\frac{\mathrm{4}!}{\mathrm{4}!\mathrm{0}!}=\frac{\mathrm{44}}{\left(\mathrm{4}!\right)^{\mathrm{11}} } \\ $$$$\frac{\frac{\mathrm{44}!}{\left(\mathrm{4}!\right)^{\mathrm{11}} }}{\mathrm{11}!}=\frac{\mathrm{44}!}{\left(\mathrm{4}!\right)^{\mathrm{11}} \mathrm{11}!} \\ $$$$\frac{\mathrm{44}!}{\left(\mathrm{4}!\right)^{\mathrm{11}} \centerdot\mathrm{11}!} \\ $$$$ \\ $$
Commented by mr W last updated on 10/Jun/25
$${thanks}\:{for}\:{trying}!\: \\ $$$${but}\:{i}\:{think}\:{the}\:{question}\:{requests}\: \\ $$$${that}\:{no}\:{pair}\:{of}\:{the}\:{pupils}\:{should}\:{be} \\ $$$${in}\:{the}\:{same}\:{group}\:{for}\:{more}\:{than} \\ $$$${one}\:{time}.\:{therefore}\:{we}\:{don}'{t}\:{have} \\ $$$${so}\:{many}\:{possibilities}. \\ $$