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Question Number 195964 by mr W last updated on 15/Aug/23

the family A has 5 members and the family B has 4 members. there are 6 personsfrom other families. in how many ways can you arrange these 15 persons around a round table such that no member from family A and no member from family B are next to each other?

$${the}\:{family}\:{A}\:{has}\:\mathrm{5}\:{members}\:{and}\:{the} \\ $$$${family}\:{B}\:{has}\:\mathrm{4}\:{members}.\:{there}\:{are}\: \\ $$$$\mathrm{6}\:{personsfrom}\:{other}\:{families}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{arrange} \\ $$$${these}\:\mathrm{15}\:{persons}\:{around}\:{a}\:{round}\:{table} \\ $$$${such}\:{that}\:{no}\:{member}\:{from}\:{family}\:{A} \\ $$$${and}\:{no}\:{member}\:{from}\:{family}\:{B}\:{are} \\ $$$${next}\:{to}\:{each}\:{other}? \\ $$

Answered by mr W last updated on 15/Aug/23

Commented by mr W last updated on 15/Aug/23

we arrange at first the 6 people from other families (green seats) on the table. there are 5! ways to do this. say ♠ stands for a seat for a member from family A and ♣ stands for a seat for a member from family B. each of the 6 places (blue boxes) between the green seats can be filled with: nothing (empty) ⇔ we add 0 to it or a seat ♠ ⇔ we add 1 to it or 2 seats ♠♠ ⇔ we add 2 to it or 3 seats ♠♠♠ ⇔ we add 3 to it or 4 seats ♠♠♠♠ ⇔ we add 4 to it or 5 seats ♠♠♠♠♠ ⇔ we add 5 to it or a seat ♣ ⇔ we add 100 to it or 2 seats ♣♣ ⇔ we add 200 to it or 3 seats ♣♣♣ ⇔ we add 300 to it or 4 seats ♣♣♣♣ ⇔ we add 400 to it in all 6 boxes we should totally put 5 seats ♠♠♠♠♠ and 4 seats ♣♣♣♣. that means the sum of all 6 boxes should be 5+400=405. so the number of ways to do this is the coefficient of term x^(405) in the expansion of the generating function (1+x+x^2 +x^3 +x^4 +x^5 +x^(100) +x^(200) +x^(300) +x^(400) )^6 , which is 4626, see below. since the members both in family A and family B are permutable, the number of ways to arrange the members from family A and B is 4626×4!×5!. the totaly number of ways to arrange all people around the table is thus 4626×4!×5!×5!=1 598 745 600

$${we}\:{arrange}\:{at}\:{first}\:{the}\:\mathrm{6}\:{people}\:{from} \\ $$$${other}\:{families}\:\left({green}\:{seats}\right)\:{on} \\ $$$${the}\:{table}.\:{there}\:{are}\:\mathrm{5}!\:{ways}\:{to}\:{do}\:{this}. \\ $$$${say}\:\spadesuit\:{stands}\:{for}\:{a}\:{seat}\:{for}\:{a}\:{member} \\ $$$${from}\:{family}\:{A}\:{and}\:\clubsuit\:{stands}\:{for}\:{a} \\ $$$${seat}\:{for}\:{a}\:{member}\:{from}\:{family}\:{B}. \\ $$$${each}\:{of}\:{the}\:\mathrm{6}\:{places}\:\left({blue}\:{boxes}\right)\: \\ $$$${between}\:{the}\:{green}\:{seats}\:{can}\:{be}\:{filled} \\ $$$${with}: \\ $$$${nothing}\:\left({empty}\right)\:\Leftrightarrow\:{we}\:{add}\:\mathrm{0}\:{to}\:{it} \\ $$$${or}\:{a}\:{seat}\:\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{1}\:{to}\:{it} \\ $$$${or}\:\mathrm{2}\:{seats}\:\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{2}\:{to}\:{it} \\ $$$${or}\:\mathrm{3}\:{seats}\:\spadesuit\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{3}\:{to}\:{it} \\ $$$${or}\:\mathrm{4}\:{seats}\:\spadesuit\spadesuit\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{4}\:{to}\:{it} \\ $$$${or}\:\mathrm{5}\:{seats}\:\spadesuit\spadesuit\spadesuit\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{5}\:{to}\:{it} \\ $$$${or}\:{a}\:{seat}\:\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{100}\:{to}\:{it} \\ $$$${or}\:\mathrm{2}\:{seats}\:\clubsuit\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{200}\:{to}\:{it} \\ $$$${or}\:\mathrm{3}\:{seats}\:\clubsuit\clubsuit\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{300}\:{to}\:{it} \\ $$$${or}\:\mathrm{4}\:{seats}\:\clubsuit\clubsuit\clubsuit\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{400}\:{to}\:{it} \\ $$$${in}\:{all}\:\mathrm{6}\:{boxes}\:{we}\:{should}\:{totally}\:{put}\: \\ $$$$\mathrm{5}\:{seats}\:\spadesuit\spadesuit\spadesuit\spadesuit\spadesuit\:{and}\:\mathrm{4}\:{seats}\:\clubsuit\clubsuit\clubsuit\clubsuit. \\ $$$${that}\:{means}\:{the}\:{sum}\:{of}\:{all}\:\mathrm{6}\:{boxes} \\ $$$${should}\:{be}\:\mathrm{5}+\mathrm{400}=\mathrm{405}.\:{so}\:{the}\:{number} \\ $$$${of}\:{ways}\:{to}\:{do}\:{this}\:{is}\:{the}\:{coefficient} \\ $$$${of}\:{term}\:{x}^{\mathrm{405}} \:{in}\:{the}\:{expansion}\:{of}\:{the} \\ $$$${generating}\:{function} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +{x}^{\mathrm{100}} +{x}^{\mathrm{200}} +{x}^{\mathrm{300}} +{x}^{\mathrm{400}} \right)^{\mathrm{6}} , \\ $$$${which}\:{is}\:\mathrm{4626},\:{see}\:{below}.\:{since}\:{the}\: \\ $$$${members}\:{both}\:{in}\:{family}\:{A}\:{and}\: \\ $$$${family}\:{B}\:{are}\:{permutable},\:{the}\:{number} \\ $$$${of}\:{ways}\:{to}\:{arrange}\:{the}\:{members}\: \\ $$$${from}\:{family}\:{A}\:{and}\:{B}\:{is}\:\mathrm{4626}×\mathrm{4}!×\mathrm{5}!. \\ $$$${the}\:{totaly}\:{number}\:{of}\:{ways}\:{to}\:{arrange} \\ $$$${all}\:{people}\:{around}\:{the}\:{table}\:{is}\:{thus} \\ $$$$\mathrm{4626}×\mathrm{4}!×\mathrm{5}!×\mathrm{5}!=\mathrm{1}\:\mathrm{598}\:\mathrm{745}\:\mathrm{600} \\ $$

Commented by mr W last updated on 15/Aug/23

Commented by MM42 last updated on 15/Aug/23

sir W how to draw geometric shapes on this page?as in the figure above

$${sir}\:\boldsymbol{{W}} \\ $$$${how}\:{to}\:{draw}\:{geometric}\:{shapes}\: \\ $$$${on}\:{this}\:{page}?{as}\:{in}\:{the}\:{figure}\:{above}\: \\ $$

Commented by mr W last updated on 15/Aug/23

try LEKH diagram.

$${try}\:{LEKH}\:{diagram}. \\ $$

Commented by MM42 last updated on 15/Aug/23

thank you so

$${thank}\:{you}\:{so} \\ $$

Commented by Tawa11 last updated on 16/Aug/23

Great work sir.

$$\mathrm{Great}\:\mathrm{work}\:\mathrm{sir}. \\ $$

Commented by York12 last updated on 16/Aug/23

sir where to learn combinatorics

$${sir}\:{where}\:{to}\:{learn}\:{combinatorics} \\ $$

Commented by York12 last updated on 16/Aug/23

or a seat ♠ ⇔ we add 1 to it or 2 seats ♠♠ ⇔ we add 2 to it or 3 seats ♠♠♠ ⇔ we add 3 to it or 4 seats ♠♠♠♠ ⇔ we add 4 to it or 5 seats ♠♠♠♠♠ ⇔ we add 5 to it or a seat ♣ ⇔ we add 100 to it or 2 seats ♣♣ ⇔ we add 200 to it or 3 seats ♣♣♣ ⇔ we add 300 to it or 4 seats ♣♣♣♣ ⇔ we add 400 to it in all 6 boxes we should totally put 5 seats ♠♠♠♠♠ and 4 seats ♣♣♣♣. that means the sum of all 6 boxes should be 5+400=405

$$ \\ $$$${or}\:{a}\:{seat}\:\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{1}\:{to}\:{it} \\ $$$${or}\:\mathrm{2}\:{seats}\:\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{2}\:{to}\:{it} \\ $$$${or}\:\mathrm{3}\:{seats}\:\spadesuit\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{3}\:{to}\:{it} \\ $$$${or}\:\mathrm{4}\:{seats}\:\spadesuit\spadesuit\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{4}\:{to}\:{it} \\ $$$${or}\:\mathrm{5}\:{seats}\:\spadesuit\spadesuit\spadesuit\spadesuit\spadesuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{5}\:{to}\:{it} \\ $$$${or}\:{a}\:{seat}\:\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{100}\:{to}\:{it} \\ $$$${or}\:\mathrm{2}\:{seats}\:\clubsuit\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{200}\:{to}\:{it} \\ $$$${or}\:\mathrm{3}\:{seats}\:\clubsuit\clubsuit\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{300}\:{to}\:{it} \\ $$$${or}\:\mathrm{4}\:{seats}\:\clubsuit\clubsuit\clubsuit\clubsuit\:\Leftrightarrow\:{we}\:{add}\:\mathrm{400}\:{to}\:{it} \\ $$$${in}\:{all}\:\mathrm{6}\:{boxes}\:{we}\:{should}\:{totally}\:{put}\: \\ $$$$\mathrm{5}\:{seats}\:\spadesuit\spadesuit\spadesuit\spadesuit\spadesuit\:{and}\:\mathrm{4}\:{seats}\:\clubsuit\clubsuit\clubsuit\clubsuit. \\ $$$${that}\:{means}\:{the}\:{sum}\:{of}\:{all}\:\mathrm{6}\:{boxes} \\ $$$${should}\:{be}\:\mathrm{5}+\mathrm{400}=\mathrm{405} \\ $$$$ \\ $$

Commented by York12 last updated on 16/Aug/23

I do not understand what do you mean by this part

$${I}\:{do}\:{not}\:{understand}\:{what}\:{do}\:{you} \\ $$$${mean}\:{by}\:{this}\:{part} \\ $$$$ \\ $$

Commented by mr W last updated on 16/Aug/23

you have 5 seats ♠for the family A. in each box you can put 1 or 2 or 3 or 4 or 5 seats ♠. totally you must put all 5 seats ♠. or you put in this only seats for family B, namly 1 or 2 or 3 or 4 seats ♣. totally you must put all 4 seats ♣. if empty means 0, a ♠ means 1, 2 ♠♠ means 2, 3 ♠♠♠ means 3, 4 ♠♠♠♠ means 4, 5 ♠♠♠♠♠ means 5, and a ♣ means 100, 2 ♣♣ means 200, 3 ♣♣♣ means 300, 4 ♣♣♣♣ means 400, the sum of numbers in all 6 boxes must be 405.

$${you}\:{have}\:\mathrm{5}\:{seats}\:\spadesuit{for}\:{the}\:{family}\:{A}. \\ $$$${in}\:{each}\:{box}\:{you}\:{can}\:{put}\:\mathrm{1}\:{or}\:\mathrm{2}\:{or}\:\mathrm{3}\:{or} \\ $$$$\mathrm{4}\:{or}\:\mathrm{5}\:{seats}\:\spadesuit.\:{totally}\:{you}\:{must}\:{put} \\ $$$${all}\:\:\mathrm{5}\:{seats}\:\spadesuit.\:{or}\:{you}\:{put}\:{in}\:{this} \\ $$$${only}\:{seats}\:{for}\:{family}\:{B},\:{namly} \\ $$$$\mathrm{1}\:{or}\:\mathrm{2}\:{or}\:\mathrm{3}\:{or}\:\mathrm{4}\:{seats}\:\clubsuit.\:{totally}\:{you} \\ $$$${must}\:{put}\:{all}\:\mathrm{4}\:{seats}\:\clubsuit. \\ $$$${if}\:{empty}\:{means}\:\mathrm{0},\:{a}\:\spadesuit\:{means}\:\mathrm{1}, \\ $$$$\mathrm{2}\:\spadesuit\spadesuit\:{means}\:\mathrm{2},\:\mathrm{3}\:\spadesuit\spadesuit\spadesuit\:{means}\:\mathrm{3},\: \\ $$$$\mathrm{4}\:\spadesuit\spadesuit\spadesuit\spadesuit\:{means}\:\mathrm{4},\:\mathrm{5}\:\spadesuit\spadesuit\spadesuit\spadesuit\spadesuit\:{means} \\ $$$$\mathrm{5},\:{and}\:{a}\:\clubsuit\:{means}\:\mathrm{100},\:\mathrm{2}\:\clubsuit\clubsuit\:{means} \\ $$$$\mathrm{200},\:\mathrm{3}\:\clubsuit\clubsuit\clubsuit\:{means}\:\mathrm{300},\:\mathrm{4}\:\clubsuit\clubsuit\clubsuit\clubsuit\: \\ $$$${means}\:\mathrm{400},\:{the}\:{sum}\:{of}\:{numbers}\:{in} \\ $$$${all}\:\mathrm{6}\:{boxes}\:{must}\:{be}\:\mathrm{405}. \\ $$

Commented by York12 last updated on 16/Aug/23

thanks

$${thanks} \\ $$$$ \\ $$

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