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In-how-many-ways-can-30-students-be-distributed-to-10-schools-if-1-each-school-should-get-at-least-one-student-2-no-restriction-




Question Number 158529 by mr W last updated on 11/Nov/21
In how many ways can 30 students  be distributed to 10 schools, if  1. each school should get at least one       student.  2. no restriction
$${In}\:{how}\:{many}\:{ways}\:{can}\:\mathrm{30}\:{students} \\ $$$${be}\:{distributed}\:{to}\:\mathrm{10}\:{schools},\:{if} \\ $$$$\mathrm{1}.\:{each}\:{school}\:{should}\:{get}\:{at}\:{least}\:{one} \\ $$$$\:\:\:\:\:{student}. \\ $$$$\mathrm{2}.\:{no}\:{restriction} \\ $$
Answered by cortano last updated on 06/Nov/21
⇒x_1 +x_2 +...+x_(10) =30 ; x_i ≥1  ⇒ (((30−1)),((10−1)) ) = (((29)),((  9)) ) =((29!)/(9! 20!))
$$\Rightarrow{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +…+{x}_{\mathrm{10}} =\mathrm{30}\:;\:{x}_{{i}} \geqslant\mathrm{1} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{30}−\mathrm{1}}\\{\mathrm{10}−\mathrm{1}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{29}}\\{\:\:\mathrm{9}}\end{pmatrix}\:=\frac{\mathrm{29}!}{\mathrm{9}!\:\mathrm{20}!} \\ $$
Commented by mr W last updated on 07/Nov/21
this is only true, when 30 (identical)  computers are distributed to 10  schools. but here the 30 students  are different objects, so the answer  is not correct.
$${this}\:{is}\:{only}\:{true},\:{when}\:\mathrm{30}\:\left({identical}\right) \\ $$$${computers}\:{are}\:{distributed}\:{to}\:\mathrm{10} \\ $$$${schools}.\:{but}\:{here}\:{the}\:\mathrm{30}\:{students} \\ $$$${are}\:{different}\:{objects},\:{so}\:{the}\:{answer} \\ $$$${is}\:{not}\:{correct}. \\ $$
Commented by ajfour last updated on 06/Nov/21
i had got    (1)  10!^(30) C_(10) ×10^(20)   (2)10^(30)
$${i}\:{had}\:{got}\:\: \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{10}!^{\mathrm{30}} {C}_{\mathrm{10}} ×\mathrm{10}^{\mathrm{20}} \\ $$$$\left(\mathrm{2}\right)\mathrm{10}^{\mathrm{30}} \\ $$$$ \\ $$
Commented by mr W last updated on 06/Nov/21
the result for (1) seems to be logic,  but it is in fact not correct, because  it is larger than the answer to (2),  and this can not be true, since  the ways with restriction must be   less than the ways without restriction.  so where is the error?
$${the}\:{result}\:{for}\:\left(\mathrm{1}\right)\:{seems}\:{to}\:{be}\:{logic}, \\ $$$${but}\:{it}\:{is}\:{in}\:{fact}\:{not}\:{correct},\:{because} \\ $$$${it}\:{is}\:{larger}\:{than}\:{the}\:{answer}\:{to}\:\left(\mathrm{2}\right), \\ $$$${and}\:{this}\:{can}\:{not}\:{be}\:{true},\:{since} \\ $$$${the}\:{ways}\:{with}\:{restriction}\:{must}\:{be}\: \\ $$$${less}\:{than}\:{the}\:{ways}\:{without}\:{restriction}. \\ $$$${so}\:{where}\:{is}\:{the}\:{error}? \\ $$
Commented by mr W last updated on 11/Nov/21
correct solution see Q158955
$${correct}\:{solution}\:{see}\:{Q}\mathrm{158955} \\ $$

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