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Question Number 21802 by Tinkutara last updated on 04/Oct/17
Five balls are to be placed in three boxes. Each can hold all the five balls. In how many different ways can we place the balls so that no box remains empty, if balls are different but boxes are identical?
$$\mathrm{Five}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{to}\:\mathrm{be}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{three} \\ $$$$\mathrm{boxes}.\:\mathrm{Each}\:\mathrm{can}\:\mathrm{hold}\:\mathrm{all}\:\mathrm{the}\:\mathrm{five}\:\mathrm{balls}. \\ $$$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{different}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{we} \\ $$$$\mathrm{place}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{so}\:\mathrm{that}\:\mathrm{no}\:\mathrm{box}\:\mathrm{remains} \\ $$$$\mathrm{empty},\:\mathrm{if}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{different}\:\mathrm{but}\:\mathrm{boxes} \\ $$$$\mathrm{are}\:\mathrm{identical}? \\ $$
Commented by mrW1 last updated on 06/Oct/17
25 ways
$$\mathrm{25}\:\mathrm{ways} \\ $$
Answered by ajfour last updated on 06/Oct/17
coeff. of x^5 in ((5!)/(3!))[x+(x^2 /(2!))+(x^3 /(3!))]^3 =coeff. of x^2 in 20(1+(x/2)+(x^2 /6))^3 =20((3/6)+(3/4)) =20×(5/4)=25 .
$${coeff}.\:{of}\:{x}^{\mathrm{5}} \:{in}\:\frac{\mathrm{5}!}{\mathrm{3}!}\left[{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\right]^{\mathrm{3}} \\ $$$$={coeff}.\:{of}\:{x}^{\mathrm{2}} \:{in}\:\mathrm{20}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{6}}\right)^{\mathrm{3}} \\ $$$$\:\:\:=\mathrm{20}\left(\frac{\mathrm{3}}{\mathrm{6}}+\frac{\mathrm{3}}{\mathrm{4}}\right)\:=\mathrm{20}×\frac{\mathrm{5}}{\mathrm{4}}=\mathrm{25}\:. \\ $$
Commented by Tinkutara last updated on 06/Oct/17
But I don′t understand why to find such coefficients in these expansions? I know this formula but don′t understand the reason.
$$\mathrm{But}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{to}\:\mathrm{find} \\ $$$$\mathrm{such}\:\mathrm{coefficients}\:\mathrm{in}\:\mathrm{these}\:\mathrm{expansions}? \\ $$$$\mathrm{I}\:\mathrm{know}\:\mathrm{this}\:\mathrm{formula}\:\mathrm{but}\:\mathrm{don}'\mathrm{t} \\ $$$$\mathrm{understand}\:\mathrm{the}\:\mathrm{reason}. \\ $$
Commented by ajfour last updated on 09/Oct/17
if x_1 +x_2 +x_3 =9 placing 9 balls in three boxes such that they can remain empty, then it is equivalent to choosing x^9 from the expansion of (1+x+x^2 +...)^3 =(1+x+x^2 +...)(1+x+x^2 +...)(1+x+x^2 +..) because powers of x from each bracket shall add and similarly no. of balls in each box shall add. the sum has to be 9 , so we choose no. of ways of obtaining x^9 = coefficient of x^9 . =coeff. of x^9 in the expansion of (1+x+x^2 +...)^3 or in (1−x)^(−3) =^(9+3−1) C_(3−1) =^(11) C_2 .
$${if}\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} =\mathrm{9} \\ $$$${placing}\:\mathrm{9}\:{balls}\:{in}\:{three}\:{boxes}\:{such} \\ $$$${that}\:{they}\:{can}\:{remain}\:{empty},\:{then} \\ $$$${it}\:{is}\:{equivalent}\:{to}\:\:{choosing}\:{x}^{\mathrm{9}} \:{from} \\ $$$${the}\:{expansion}\:{of} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +..\right) \\ $$$${because}\:{powers}\:{of}\:{x}\:{from}\:{each} \\ $$$${bracket}\:{shall}\:{add}\:{and}\:{similarly} \\ $$$${no}.\:{of}\:{balls}\:{in}\:{each}\:{box}\:{shall}\:{add}. \\ $$$${the}\:{sum}\:{has}\:{to}\:{be}\:\mathrm{9}\:,\:{so}\:{we}\:{choose} \\ $$$${no}.\:{of}\:{ways}\:{of}\:{obtaining}\:{x}^{\mathrm{9}} \:= \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{9}} \:. \\ $$$$={coeff}.\:{of}\:{x}^{\mathrm{9}} \:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{3}} \\ $$$${or}\:{in}\:\left(\mathrm{1}−{x}\right)^{−\mathrm{3}} \:=\:^{\mathrm{9}+\mathrm{3}−\mathrm{1}} {C}_{\mathrm{3}−\mathrm{1}} =^{\mathrm{11}} {C}_{\mathrm{2}} \:. \\ $$
Commented by mrW1 last updated on 06/Oct/17
nice explained! as result one can use the notation of stirling number {_3 ^5 } which means the number of ways to put 5 different objects into 3 identical boxes. {_3 ^5 }=25
$$\mathrm{nice}\:\mathrm{explained}! \\ $$$$ \\ $$$$\mathrm{as}\:\mathrm{result}\:\mathrm{one}\:\mathrm{can}\:\mathrm{use}\:\mathrm{the}\:\mathrm{notation}\:\mathrm{of} \\ $$$$\mathrm{stirling}\:\mathrm{number}\:\left\{_{\mathrm{3}} ^{\mathrm{5}} \right\}\:\mathrm{which}\:\mathrm{means} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{put}\:\mathrm{5}\:\mathrm{different} \\ $$$$\mathrm{objects}\:\mathrm{into}\:\mathrm{3}\:\mathrm{identical}\:\mathrm{boxes}. \\ $$$$\left\{_{\mathrm{3}} ^{\mathrm{5}} \right\}=\mathrm{25} \\ $$
Commented by Tinkutara last updated on 06/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by mrW1 last updated on 09/Oct/17
to mr. ajfour: you said: =coeff. of x^9 in the expansion of (1+x+x^2 +...)^3 or in (1−x)^(−3) =^(9−3+1) C_3 =^7 C_3 . should the last line not be: or in (1−x)^(−3) =^(9+3−1) C_9 =^(11) C_9 . ?
$$\mathrm{to}\:\mathrm{mr}.\:\mathrm{ajfour}: \\ $$$$\mathrm{you}\:\mathrm{said}: \\ $$$$={coeff}.\:{of}\:{x}^{\mathrm{9}} \:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{3}} \\ $$$${or}\:{in}\:\left(\mathrm{1}−{x}\right)^{−\mathrm{3}} \:=\:^{\mathrm{9}−\mathrm{3}+\mathrm{1}} {C}_{\mathrm{3}} =^{\mathrm{7}} {C}_{\mathrm{3}} \:. \\ $$$$ \\ $$$$\mathrm{should}\:\mathrm{the}\:\mathrm{last}\:\mathrm{line}\:\mathrm{not}\:\mathrm{be}: \\ $$$${or}\:{in}\:\left(\mathrm{1}−{x}\right)^{−\mathrm{3}} \:=\:^{\mathrm{9}+\mathrm{3}−\mathrm{1}} {C}_{\mathrm{9}} =^{\mathrm{11}} {C}_{\mathrm{9}} \:. \\ $$$$? \\ $$
Commented by ajfour last updated on 09/Oct/17
yes it is in fact as you say.. i dint remember correctly, n thought i did.
$${yes}\:{it}\:{is}\:{in}\:{fact}\:{as}\:{you}\:{say}.. \\ $$$${i}\:{dint}\:{remember}\:{correctly},\:{n}\: \\ $$$${thought}\:{i}\:{did}. \\ $$

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