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Question-133642

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Question Number 133642 by liberty last updated on 23/Feb/21
Commented by liberty last updated on 23/Feb/21
dear Mr W, Mr Edwin and all master
$$\mathrm{dear}\:\mathrm{Mr}\:\mathrm{W},\:\mathrm{Mr}\:\mathrm{Edwin}\:\mathrm{and}\:\mathrm{all}\:\mathrm{master} \\ $$
Answered by bobhans last updated on 23/Feb/21
It′s the same as the number of ways of placing 2 identical dividers in a line of 12 identical 12 balls ,which is the same as the number of ways of choosing 2 items from 14 distinct items = C _2^(14) = 91
$${It}'{s}\:{the}\:{same}\:{as}\:{the}\:{number}\:{of}\:{ways} \\ $$$${of}\:{placing}\:\mathrm{2}\:{identical}\:{dividers}\:{in} \\ $$$${a}\:{line}\:{of}\:\mathrm{12}\:{identical}\:\mathrm{12}\:{balls}\:,{which} \\ $$$${is}\:{the}\:{same}\:{as}\:{the}\:{number}\:{of}\:{ways} \\ $$$${of}\:{choosing}\:\mathrm{2}\:{items}\:{from}\:\mathrm{14}\: \\ $$$${distinct}\:{items}\:=\:{C}\:_{\mathrm{2}} ^{\mathrm{14}} \:=\:\mathrm{91} \\ $$
Commented by liberty last updated on 23/Feb/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 23/Feb/21
□∣□∣□∣□∣□∣□∣□∣□∣□∣□∣□∣□ if no box is empty, then the answer is C_(3−1) ^(12−1) =C_2 ^(11) =55
$$\square\mid\square\mid\square\mid\square\mid\square\mid\square\mid\square\mid\square\mid\square\mid\square\mid\square\mid\square \\ $$$${if}\:{no}\:{box}\:{is}\:{empty},\:{then}\:{the}\:{answer}\:{is} \\ $$$${C}_{\mathrm{3}−\mathrm{1}} ^{\mathrm{12}−\mathrm{1}} ={C}_{\mathrm{2}} ^{\mathrm{11}} =\mathrm{55} \\ $$
Commented by liberty last updated on 23/Feb/21
how you got C_2 ^(11) sir? let the boxes namely A,B and C (1,1,10),(1,2,9),(1,3,8),(1,4,7) (1,5,6),(2,2,8),(2,3,7)...
$$\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{11}} \:\mathrm{sir}? \\ $$$$\mathrm{let}\:\mathrm{the}\:\mathrm{boxes}\:\mathrm{namely}\:\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{C} \\ $$$$\left(\mathrm{1},\mathrm{1},\mathrm{10}\right),\left(\mathrm{1},\mathrm{2},\mathrm{9}\right),\left(\mathrm{1},\mathrm{3},\mathrm{8}\right),\left(\mathrm{1},\mathrm{4},\mathrm{7}\right) \\ $$$$\left(\mathrm{1},\mathrm{5},\mathrm{6}\right),\left(\mathrm{2},\mathrm{2},\mathrm{8}\right),\left(\mathrm{2},\mathrm{3},\mathrm{7}\right)… \\ $$
Commented by mr W last updated on 23/Feb/21
as shown above, we have 12 balls □, to divide them into 3 groups we only need to place 2 bars ∣ among them. there are 11 possible positions ∣ to place these 2 bars, so the answer is C_2 ^(11) . this method is called “stars and bars” method.
$${as}\:{shown}\:{above},\:{we}\:{have}\:\mathrm{12}\:{balls}\:\square,\:{to} \\ $$$${divide}\:{them}\:{into}\:\mathrm{3}\:{groups}\:{we}\:{only} \\ $$$${need}\:{to}\:{place}\:\mathrm{2}\:{bars}\:\mid\:{among}\:{them}. \\ $$$${there}\:{are}\:\mathrm{11}\:{possible}\:{positions}\:\mid\:{to} \\ $$$${place}\:{these}\:\mathrm{2}\:{bars},\:{so}\:{the}\:{answer}\:{is} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{11}} .\:{this}\:{method}\:{is}\:{called}\:“{stars}\:{and} \\ $$$${bars}''\:{method}. \\ $$
Commented by mr W last updated on 23/Feb/21
or an other way: (x+x^2 +x^3 +...)^3 =(x^3 /((1−x)^3 ))=x^3 Σ_(k=0) ^∞ C_2 ^(k+2) x^k coef. of x^(12) is the answer =C_2 ^(9+2) =C_2 ^(11) if the boxes may be empty, then (1+x+x^2 +x^3 +...)^3 =(1/((1−x)^3 ))=Σ_(k=0) ^∞ C_2 ^(k+2) x^k coef. of x^(12) is the answer =C_2 ^(12+2) =C_2 ^(14) =91
$${or}\:{an}\:{other}\:{way}: \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{3}} =\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }={x}^{\mathrm{3}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{12}} \:{is}\:{the}\:{answer}\:={C}_{\mathrm{2}} ^{\mathrm{9}+\mathrm{2}} ={C}_{\mathrm{2}} ^{\mathrm{11}} \\ $$$$ \\ $$$${if}\:{the}\:{boxes}\:{may}\:{be}\:{empty},\:{then} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{12}} \:{is}\:{the}\:{answer}\:={C}_{\mathrm{2}} ^{\mathrm{12}+\mathrm{2}} ={C}_{\mathrm{2}} ^{\mathrm{14}} =\mathrm{91} \\ $$
Commented by liberty last updated on 23/Feb/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mr W last updated on 23/Feb/21
but to divide 12 distinct balls into 3 distinct boxes there are 519156 ways!
$${but}\:{to}\:{divide}\:\mathrm{12}\:{distinct}\:{balls}\:{into}\:\mathrm{3} \\ $$$${distinct}\:{boxes}\:{there}\:{are}\:\mathrm{519156}\:{ways}! \\ $$
Commented by liberty last updated on 23/Feb/21
how to find it?
$$\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{it}? \\ $$
Commented by mr W last updated on 23/Feb/21
using generating function it is the coefficient of x^(12) term in the expansion of 12!(e^x −1)^3 .
$${using}\:{generating}\:{function}\:{it}\:{is}\:{the} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{12}} \:{term}\:{in}\:{the} \\ $$$${expansion}\:{of}\:\mathrm{12}!\left({e}^{{x}} −\mathrm{1}\right)^{\mathrm{3}} . \\ $$
Answered by EDWIN88 last updated on 23/Feb/21
the question is equivalent to how many positive number integer solution a+b+c = 12 = C_2 ^( 11) = 55 but if the equation how many non negatif number (whole number ) are there a+b+c=100 = C_2 ^( 14) = 91
$$\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{how}\:\mathrm{many}\: \\ $$$$\mathrm{positive}\:\mathrm{number}\:\mathrm{integer}\:\mathrm{solution}\:{a}+{b}+{c}\:=\:\mathrm{12} \\ $$$$=\:{C}_{\mathrm{2}} ^{\:\mathrm{11}} \:=\:\mathrm{55}\: \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{how}\:\mathrm{many}\:\mathrm{non}\:\mathrm{negatif}\:\mathrm{number} \\ $$$$\left(\mathrm{whole}\:\mathrm{number}\:\right)\:\mathrm{are}\:\mathrm{there}\:{a}+{b}+{c}=\mathrm{100} \\ $$$$=\:{C}_{\mathrm{2}} ^{\:\mathrm{14}} \:=\:\mathrm{91} \\ $$
Commented by liberty last updated on 23/Feb/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by JDamian last updated on 23/Feb/21
This can be thought as the number of Permutations with repetition of 12 balls + (3−1) separators •••◊••••◊••••• (((12+3−1)!)/(12! ∙ (3−1)!)) = ((14!)/(12! ∙ 2!)) = C_2 ^(14)
$${This}\:{can}\:{be}\:{thought}\:{as}\:{the}\:{number}\:{of} \\ $$$${Permutations}\:{with}\:{repetition}\:{of} \\ $$$$\mathrm{12}\:{balls}\:+\:\left(\mathrm{3}−\mathrm{1}\right)\:{separators} \\ $$$$ \\ $$$$\bullet\bullet\bullet\lozenge\bullet\bullet\bullet\bullet\lozenge\bullet\bullet\bullet\bullet\bullet \\ $$$$ \\ $$$$\frac{\left(\mathrm{12}+\mathrm{3}−\mathrm{1}\right)!}{\mathrm{12}!\:\centerdot\:\left(\mathrm{3}−\mathrm{1}\right)!}\:=\:\frac{\mathrm{14}!}{\mathrm{12}!\:\centerdot\:\mathrm{2}!}\:=\:{C}_{\mathrm{2}} ^{\mathrm{14}} \\ $$

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