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Question Number 198851 by Tawa11 last updated on 25/Oct/23
You want to arrange 17 books on the shelf of a bookstore.  The shelf is dedicated to the three Toni Morrison novels  published between 1977 and 1987: Song of Solomon,  Tar Baby, and Beloved. You have many copies of each,  but on the shelf you want an even number of Song of  Solomon, at least three copies of Tar Baby, and at most  four copies of Beloved. How many different arrangements  are possible?
You want to arrange 17 books on the shelf of a bookstore.
The shelf is dedicated to the three Toni Morrison novels
published between 1977 and 1987: Song of Solomon,
Tar Baby, and Beloved. You have many copies of each,
but on the shelf you want an even number of Song of
Solomon, at least three copies of Tar Baby, and at most
four copies of Beloved. How many different arrangements
are possible?
Commented by mr W last updated on 25/Oct/23
do you have the answer?
$${do}\:{you}\:{have}\:{the}\:{answer}? \\ $$
Commented by mr W last updated on 25/Oct/23
nikif99 sir: can you use your fortran  program to calculate this?
$${nikif}\mathrm{99}\:{sir}:\:{can}\:{you}\:{use}\:{your}\:{fortran} \\ $$$${program}\:{to}\:{calculate}\:{this}? \\ $$
Commented by Tawa11 last updated on 25/Oct/23
yes sir
$$\mathrm{yes}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 25/Oct/23
what′s the answer given? is there  also a solution given?  i′m interested in alternative solutions.
$${what}'{s}\:{the}\:{answer}\:{given}?\:{is}\:{there} \\ $$$${also}\:{a}\:{solution}\:{given}? \\ $$$${i}'{m}\:{interested}\:{in}\:{alternative}\:{solutions}. \\ $$
Commented by Tawa11 last updated on 25/Oct/23
18,072,242
$$\mathrm{18},\mathrm{072},\mathrm{242} \\ $$
Commented by Tawa11 last updated on 25/Oct/23
yes sir
$$\mathrm{yes}\:\mathrm{sir} \\ $$
Commented by Tawa11 last updated on 25/Oct/23
Should I send the solution now sir?
$$\mathrm{Should}\:\mathrm{I}\:\mathrm{send}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{now}\:\mathrm{sir}? \\ $$
Commented by nikif99 last updated on 25/Oct/23
if 0 books are allowed for “Song of   Solomon” or “Beloved” I found  18.072.242 cases.  If not, then cases are limited to  18.003.510.  Left 3 columns are possible number of  copies selected of three titles, when  0 books are allowed.  Right 3 columns excempt 0 books.  For each horizontal line, number of  arrangements with repetitions is  ((17!)/(a!×b!×c!))
$${if}\:\mathrm{0}\:{books}\:{are}\:{allowed}\:{for}\:“{Song}\:{of}\: \\ $$$${Solomon}''\:{or}\:“{Beloved}''\:{I}\:{found} \\ $$$$\mathrm{18}.\mathrm{072}.\mathrm{242}\:{cases}. \\ $$$${If}\:{not},\:{then}\:{cases}\:{are}\:{limited}\:{to} \\ $$$$\mathrm{18}.\mathrm{003}.\mathrm{510}. \\ $$$${Left}\:\mathrm{3}\:{columns}\:{are}\:{possible}\:{number}\:{of} \\ $$$${copies}\:{selected}\:{of}\:{three}\:{titles},\:{when} \\ $$$$\mathrm{0}\:{books}\:{are}\:{allowed}. \\ $$$${Right}\:\mathrm{3}\:{columns}\:{excempt}\:\mathrm{0}\:{books}. \\ $$$${For}\:{each}\:{horizontal}\:{line},\:{number}\:{of} \\ $$$${arrangements}\:{with}\:{repetitions}\:{is} \\ $$$$\frac{\mathrm{17}!}{{a}!×{b}!×{c}!} \\ $$$$ \\ $$
Commented by nikif99 last updated on 25/Oct/23
Commented by Tawa11 last updated on 25/Oct/23
Commented by mr W last updated on 25/Oct/23
are you also familiar with generating  functions?
$${are}\:{you}\:{also}\:{familiar}\:{with}\:{generating} \\ $$$${functions}? \\ $$
Commented by Tawa11 last updated on 25/Oct/23
No sir.  I saw the method in your previous solution so,  I bring the question here.
$$\mathrm{No}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{saw}\:\mathrm{the}\:\mathrm{method}\:\mathrm{in}\:\mathrm{your}\:\mathrm{previous}\:\mathrm{solution}\:\mathrm{so}, \\ $$$$\mathrm{I}\:\mathrm{bring}\:\mathrm{the}\:\mathrm{question}\:\mathrm{here}. \\ $$
Answered by mr W last updated on 26/Oct/23
once again i′ll try to use generating  functions.  s=number of books “Song of Solomon”  t=number of books “Tar Baby”  b=number of books “Beloved”  s⇒(x^2 +x^4 +x^6 +...)  t⇒(x^3 +x^4 +x^5 +...)  b⇒(x+x^2 +x^3 +x^4 )  s+t+b=17  number of arrangements is the  coefficient of x^(17)  term in expansion  17!((x^2 /(2!))+(x^4 /(4!))+(x^6 /(6!))+...)((x^3 /(3!))+(x^4 /(4!))+(x^5 /(5!))+...)(x+(x^2 /(2!))+(x^3 /(3!))+(x^4 /(4!)))  which is 18 003 510.     if no book from S anf B is also  allowed, then the answer is the  coefficient of x^(17)  term in  17!(1+(x^2 /(2!))+(x^4 /(4!))+(x^6 /(6!))+...)((x^3 /(3!))+(x^4 /(4!))+(x^5 /(5!))+...)(1+x+(x^2 /(2!))+(x^3 /(3!))+(x^4 /(4!)))  which is 18 072 242.
$${once}\:{again}\:{i}'{ll}\:{try}\:{to}\:{use}\:{generating} \\ $$$${functions}. \\ $$$${s}={number}\:{of}\:{books}\:“{Song}\:{of}\:{Solomon}'' \\ $$$${t}={number}\:{of}\:{books}\:“{Tar}\:{Baby}'' \\ $$$${b}={number}\:{of}\:{books}\:“{Beloved}'' \\ $$$${s}\Rightarrow\left({x}^{\mathrm{2}} +{x}^{\mathrm{4}} +{x}^{\mathrm{6}} +…\right) \\ $$$${t}\Rightarrow\left({x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +…\right) \\ $$$${b}\Rightarrow\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} \right) \\ $$$${s}+{t}+{b}=\mathrm{17} \\ $$$${number}\:{of}\:{arrangements}\:{is}\:{the} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{17}} \:{term}\:{in}\:{expansion} \\ $$$$\mathrm{17}!\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+…\right)\left(\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+…\right)\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}\right) \\ $$$${which}\:{is}\:\mathrm{18}\:\mathrm{003}\:\mathrm{510}.\: \\ $$$$ \\ $$$${if}\:{no}\:{book}\:{from}\:{S}\:{anf}\:{B}\:{is}\:{also} \\ $$$${allowed},\:{then}\:{the}\:{answer}\:{is}\:{the} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{17}} \:{term}\:{in} \\ $$$$\mathrm{17}!\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+…\right)\left(\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+…\right)\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}\right) \\ $$$${which}\:{is}\:\mathrm{18}\:\mathrm{072}\:\mathrm{242}. \\ $$
Commented by mr W last updated on 25/Oct/23
Commented by mr W last updated on 25/Oct/23

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