Question Number 212686 by mr W last updated on 21/Oct/24
$${in}\:{how}\:{many}\:{ways}\:{can}\:{a}\:{teacher} \\ $$$${divide}\:{his}\:\mathrm{10}\:{studens}\:{into}\:\mathrm{4}\:{groups} \\ $$$${such}\:{that}\:{each}\:{group}\:{has}\:{at}\:{least}\:\mathrm{2}\: \\ $$$${students}? \\ $$
Commented by Spillover last updated on 21/Oct/24
$$\mathrm{2268000}? \\ $$
Commented by mr W last updated on 21/Oct/24
$${how}? \\ $$
Commented by Spillover last updated on 21/Oct/24
$${is}\:{it}\:{correct}? \\ $$
Commented by Spillover last updated on 21/Oct/24
$$\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{4}} }=\mathrm{2268000} \\ $$
Answered by mehdee7396 last updated on 21/Oct/24
$$\mathrm{2}/\mathrm{2}/\mathrm{2}/\mathrm{4}\:\:{or}\:\:\mathrm{2}/\mathrm{2}/\mathrm{3}/\mathrm{3} \\ $$$$\Rightarrow\begin{pmatrix}{\:\:\:\:\mathrm{10}}\\{\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4}}\end{pmatrix}+\begin{pmatrix}{\:\:\:\:\mathrm{10}}\\{\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{3}}\end{pmatrix}=\:\mathrm{44100}\: \\ $$$$ \\ $$
Commented by mr W last updated on 21/Oct/24
$${your}\:{answer}\:{were}\:{correct}\:{if}\:{the} \\ $$$${question}\:{is}\:{to}\:{divide}\:{the}\:{students} \\ $$$${into}\:\mathrm{4}\:{groups}\:{for}\:{mathmatics},\: \\ $$$${physics},\:{chemistry}\:{and}\:{biology}\: \\ $$$${respectively},\:{i}.{e}.\:{distinct}\:{objects} \\ $$$${into}\:{distinct}\:{bins}.\:{but}\:{in}\:{this}\:{question} \\ $$$${we}\:{just}\:{divide}\:\mathrm{10}\:{students}\:{into}\:\mathrm{4} \\ $$$$\left({identical}\right)\:{groups}. \\ $$
Commented by mehdee7396 last updated on 21/Oct/24
$${you}\:{are}\:{right}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by mr W last updated on 21/Oct/24
$$\mathrm{2}/\mathrm{2}/\mathrm{2}/\mathrm{4}\:\Rightarrow\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!\mathrm{4}!}=\mathrm{3150} \\ $$$$\mathrm{2}/\mathrm{2}/\mathrm{3}/\mathrm{3}\:\Rightarrow\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!\left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!}=\mathrm{6300} \\ $$$$\Sigma:\:\mathrm{3150}+\mathrm{6300}=\mathrm{9450} \\ $$
Answered by Spillover last updated on 21/Oct/24
$${Student}=\mathrm{10}\left({n}\right)\:{say} \\ $$$${group}=\mathrm{4}\left({k}\right)\:\:\:{say} \\ $$$${selected}\:{student}\:{in}\:{each}\:{group}=\left(\mathrm{2}×\mathrm{4}\right)=\mathrm{8} \\ $$$$\mathrm{10}−\mathrm{8}=\mathrm{2} \\ $$$${Number}\:{of}\:{arrangement}=\begin{pmatrix}{{n}+{k}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{n}+{k}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}+\mathrm{4}−\mathrm{1}}\\{\mathrm{4}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{5}}\\{\mathrm{3}}\end{pmatrix}=\mathrm{10} \\ $$$${Total}\:{way}=\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{4}} }×\mathrm{10}=\mathrm{2268000} \\ $$
Commented by mr W last updated on 21/Oct/24
$${this}\:{is}\:{wrong}\:{sir}. \\ $$$${to}\:{divide}\:\mathrm{10}\:{distinct}\:{objects}\:{into}\:\mathrm{4} \\ $$$${identical}\:{bins}\:{such}\:{that}\:{no}\:{bin}\:{is} \\ $$$${empty},\:{there}\:{are}\:\left\{_{\mathrm{4}} ^{\mathrm{10}} \right\}={S}\left(\mathrm{10},\mathrm{4}\right)=\mathrm{34105} \\ $$$${ways}.\:{since}\:{in}\:{our}\:{question}\:{each}\:{bin} \\ $$$${should}\:{have}\:{at}\:{least}\:\mathrm{2}\:{objects},\:{so}\:{the} \\ $$$${answer}\:{should}\:{be}\:{less}\:{than}\:\mathrm{34105}. \\ $$
Commented by mr W last updated on 21/Oct/24
$$\left\{_{\mathrm{4}} ^{\mathrm{10}} \right\}\:{or}\:{S}\left(\mathrm{10},\mathrm{4}\right)\:{are}\:{the}\:{stirling} \\ $$$${numbers}\:{of}\:{the}\:{second}\:{kind}. \\ $$
Answered by golsendro last updated on 21/Oct/24
$$\left(\mathrm{1}\right)\:\frac{\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{4}}\\{\mathrm{4}}\end{pmatrix}}{\mathrm{3}!} \\ $$$$\:\left(\mathrm{2}\right)\:\frac{\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}}{\left(\mathrm{2}!\right)^{\mathrm{4}} } \\ $$
Commented by mr W last updated on 21/Oct/24
$${you}\:{mean} \\ $$$$\frac{\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}}{\left(\mathrm{2}!\right)^{\mathrm{2}} }\:? \\ $$