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Question Number 212686 by mr W last updated on 21/Oct/24
in how many ways can a teacher divide his 10 studens into 4 groups such that each group has at least 2 students?
$${in}\:{how}\:{many}\:{ways}\:{can}\:{a}\:{teacher} \\ $$$${divide}\:{his}\:\mathrm{10}\:{studens}\:{into}\:\mathrm{4}\:{groups} \\ $$$${such}\:{that}\:{each}\:{group}\:{has}\:{at}\:{least}\:\mathrm{2}\: \\ $$$${students}? \\ $$
Commented by Spillover last updated on 21/Oct/24
2268000?
$$\mathrm{2268000}? \\ $$
Commented by mr W last updated on 21/Oct/24
how?
$${how}? \\ $$
Commented by Spillover last updated on 21/Oct/24
is it correct?
$${is}\:{it}\:{correct}? \\ $$
Commented by Spillover last updated on 21/Oct/24
((10!)/((2!)^4 ))=2268000
$$\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{4}} }=\mathrm{2268000} \\ $$
Answered by mehdee7396 last updated on 21/Oct/24
2/2/2/4 or 2/2/3/3 ⇒ ((( 10)),((2,2,2,4)) )+ ((( 10)),((2,2,3,3)) )= 44100
$$\mathrm{2}/\mathrm{2}/\mathrm{2}/\mathrm{4}\:\:{or}\:\:\mathrm{2}/\mathrm{2}/\mathrm{3}/\mathrm{3} \\ $$$$\Rightarrow\begin{pmatrix}{\:\:\:\:\mathrm{10}}\\{\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4}}\end{pmatrix}+\begin{pmatrix}{\:\:\:\:\mathrm{10}}\\{\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{3}}\end{pmatrix}=\:\mathrm{44100}\: \\ $$$$ \\ $$
Commented by mr W last updated on 21/Oct/24
your answer were correct if the question is to divide the students into 4 groups for mathmatics, physics, chemistry and biology respectively, i.e. distinct objects into distinct bins. but in this question we just divide 10 students into 4 (identical) groups.
$${your}\:{answer}\:{were}\:{correct}\:{if}\:{the} \\ $$$${question}\:{is}\:{to}\:{divide}\:{the}\:{students} \\ $$$${into}\:\mathrm{4}\:{groups}\:{for}\:{mathmatics},\: \\ $$$${physics},\:{chemistry}\:{and}\:{biology}\: \\ $$$${respectively},\:{i}.{e}.\:{distinct}\:{objects} \\ $$$${into}\:{distinct}\:{bins}.\:{but}\:{in}\:{this}\:{question} \\ $$$${we}\:{just}\:{divide}\:\mathrm{10}\:{students}\:{into}\:\mathrm{4} \\ $$$$\left({identical}\right)\:{groups}. \\ $$
Commented by mehdee7396 last updated on 21/Oct/24
you are right ⋛
$${you}\:{are}\:{right}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by mr W last updated on 21/Oct/24
2/2/2/4 ⇒((10!)/((2!)^3 3!4!))=3150 2/2/3/3 ⇒((10!)/((2!)^2 2!(3!)^2 2!))=6300 Σ: 3150+6300=9450
$$\mathrm{2}/\mathrm{2}/\mathrm{2}/\mathrm{4}\:\Rightarrow\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{3}} \mathrm{3}!\mathrm{4}!}=\mathrm{3150} \\ $$$$\mathrm{2}/\mathrm{2}/\mathrm{3}/\mathrm{3}\:\Rightarrow\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{2}} \mathrm{2}!\left(\mathrm{3}!\right)^{\mathrm{2}} \mathrm{2}!}=\mathrm{6300} \\ $$$$\Sigma:\:\mathrm{3150}+\mathrm{6300}=\mathrm{9450} \\ $$
Answered by Spillover last updated on 21/Oct/24
Student=10(n) say group=4(k) say selected student in each group=(2×4)=8 10−8=2 Number of arrangement= (((n+k−1)),((k−1)) ) (((n+k−1)),((k−1)) )= (((2+4−1)),((4−1)) )= ((5),(3) )=10 Total way=((10!)/((2!)^4 ))×10=2268000
$${Student}=\mathrm{10}\left({n}\right)\:{say} \\ $$$${group}=\mathrm{4}\left({k}\right)\:\:\:{say} \\ $$$${selected}\:{student}\:{in}\:{each}\:{group}=\left(\mathrm{2}×\mathrm{4}\right)=\mathrm{8} \\ $$$$\mathrm{10}−\mathrm{8}=\mathrm{2} \\ $$$${Number}\:{of}\:{arrangement}=\begin{pmatrix}{{n}+{k}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{n}+{k}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{2}+\mathrm{4}−\mathrm{1}}\\{\mathrm{4}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{5}}\\{\mathrm{3}}\end{pmatrix}=\mathrm{10} \\ $$$${Total}\:{way}=\frac{\mathrm{10}!}{\left(\mathrm{2}!\right)^{\mathrm{4}} }×\mathrm{10}=\mathrm{2268000} \\ $$
Commented by mr W last updated on 21/Oct/24
this is wrong sir. to divide 10 distinct objects into 4 identical bins such that no bin is empty, there are {_4 ^(10) }=S(10,4)=34105 ways. since in our question each bin should have at least 2 objects, so the answer should be less than 34105.
$${this}\:{is}\:{wrong}\:{sir}. \\ $$$${to}\:{divide}\:\mathrm{10}\:{distinct}\:{objects}\:{into}\:\mathrm{4} \\ $$$${identical}\:{bins}\:{such}\:{that}\:{no}\:{bin}\:{is} \\ $$$${empty},\:{there}\:{are}\:\left\{_{\mathrm{4}} ^{\mathrm{10}} \right\}={S}\left(\mathrm{10},\mathrm{4}\right)=\mathrm{34105} \\ $$$${ways}.\:{since}\:{in}\:{our}\:{question}\:{each}\:{bin} \\ $$$${should}\:{have}\:{at}\:{least}\:\mathrm{2}\:{objects},\:{so}\:{the} \\ $$$${answer}\:{should}\:{be}\:{less}\:{than}\:\mathrm{34105}. \\ $$
Commented by mr W last updated on 22/Oct/24
{_4 ^(10) } or S(10,4) is the stirling number of the second kind.
$$\left\{_{\mathrm{4}} ^{\mathrm{10}} \right\}\:{or}\:{S}\left(\mathrm{10},\mathrm{4}\right)\:{is}\:{the}\:{stirling} \\ $$$${number}\:{of}\:{the}\:{second}\:{kind}. \\ $$
Answered by golsendro last updated on 22/Oct/24
(1) (( (((10)),(( 2)) ) ((8),(2) ) ((6),(2) ) ((4),(4) ))/(3!)) (2) (( (((10)),(( 2)) ) ((8),(2) ) ((6),(3) ) ((3),(3) ))/((2!)^2 ))
$$\left(\mathrm{1}\right)\:\frac{\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{4}}\\{\mathrm{4}}\end{pmatrix}}{\mathrm{3}!} \\ $$$$\:\left(\mathrm{2}\right)\:\frac{\begin{pmatrix}{\mathrm{10}}\\{\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}}{\left(\mathrm{2}!\right)^{\mathrm{2}} } \\ $$
Commented by golsendro last updated on 22/Oct/24
yes
$$\mathrm{yes} \\ $$

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