Question-95723

Question Number 95723 by i jagooll last updated on 27/May/20

Commented by bobhans last updated on 27/May/20

the case is ssma with there are 5 women 2 men sitting in 7 chairs with no 2 men sitting side by side

Answered by Rio Michael last updated on 27/May/20

the 5 girls can stand in a row in 5! ways let the girls be labeled as below G_1 G_2 G_3 G_4 G_5 the first boy coming in has 6 vacancies the second has 5 vacanciies the third has 4 vacancies so ⇒ number of ways = 5! × 6 × 5× 4 = 14 400 ways

$$\mathrm{the}\:\mathrm{5}\:\mathrm{girls}\:\mathrm{can}\:\mathrm{stand}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row}\:\mathrm{in}\:\mathrm{5}!\:\mathrm{ways} \\ $$$$\mathrm{let}\:\mathrm{the}\:\mathrm{girls}\:\mathrm{be}\:\mathrm{labeled}\:\mathrm{as}\:\mathrm{below} \\ $$$$\:{G}_{\mathrm{1}} \:\:\:\:\:{G}_{\mathrm{2}} \:\:{G}_{\mathrm{3}} \:\:{G}_{\mathrm{4}} \:\:\:{G}_{\mathrm{5}} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{boy}\:\mathrm{coming}\:\mathrm{in}\:\mathrm{has}\:\mathrm{6}\:\mathrm{vacancies} \\ $$$$\mathrm{the}\:\mathrm{second}\:\mathrm{has}\:\mathrm{5}\:\mathrm{vacanciies} \\ $$$$\mathrm{the}\:\mathrm{third}\:\mathrm{has}\:\mathrm{4}\:\mathrm{vacancies}\:\mathrm{so}\: \\ $$$$\Rightarrow\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:=\:\mathrm{5}!\:×\:\mathrm{6}\:×\:\mathrm{5}×\:\mathrm{4}\:=\:\mathrm{14}\:\mathrm{400}\:\mathrm{ways} \\ $$

Commented by mr W last updated on 27/May/20

$$\mathrm{14400}\:{is}\:{correct}!\:{Rio}\:{Michael}\:{sir}: \\ $$$${where}\:{do}\:{you}\:{think}\:{you}\:{are}\:{wrong}? \\ $$

Commented by i jagooll last updated on 27/May/20

$$\mathrm{your}\:\mathrm{answer}\:\mathrm{not}\:\mathrm{include}\:\mathrm{G}\:\mathrm{G}\:\mathrm{BGBGGB}? \\ $$

Commented by Rio Michael last updated on 28/May/20

well sir i was in a hurry on that problem when the others dropped thier solution i thought i misread it.

$$\mathrm{well}\:\mathrm{sir}\:\mathrm{i}\:\mathrm{was}\:\mathrm{in}\:\mathrm{a}\:\mathrm{hurry}\:\mathrm{on}\:\mathrm{that}\:\mathrm{problem} \\ $$$$\mathrm{when}\:\mathrm{the}\:\mathrm{others}\:\mathrm{dropped}\:\mathrm{thier}\:\mathrm{solution} \\ $$$$\mathrm{i}\:\mathrm{thought}\:\mathrm{i}\:\mathrm{misread}\:\mathrm{it}. \\ $$

Answered by bobhans last updated on 27/May/20

totally arrangement = 8! 2 boy stand are together = C_2 ^3 .3! 6! then = 8! − 6.6! = 6! ×50

$$\mathrm{totally}\:\mathrm{arrangement}\:=\:\mathrm{8}! \\ $$$$\mathrm{2}\:\mathrm{boy}\:\mathrm{stand}\:\mathrm{are}\:\mathrm{together}\:=\:\mathrm{C}_{\mathrm{2}} ^{\mathrm{3}} .\mathrm{3}!\:\mathrm{6}! \\ $$$$\mathrm{then}\:=\:\mathrm{8}!\:−\:\mathrm{6}.\mathrm{6}!\:=\:\mathrm{6}!\:×\mathrm{50}\: \\ $$

Answered by mr W last updated on 27/May/20

METHOD I between two boys there must be at least one girl. we arrange at first the 5 girls, there are 5! ways. □G□G□G□G□G□ the 3 boys can now take three of 6 positions (□). there are C_3 ^6 ×3! ways. so totally there are C_3 ^6 ×3!×5!=14400 ways. METHOD II to arrange 8 children there are 8! ways. to arrange them such that three boys are together, there are 6!×3! ways. to arrange them such that two and only two boys are next to each other, there are C_2 ^3 ×(7!−2×6!)×2! ways. total valid arrangements: 8!−C_2 ^3 ×(7!−2×6!)×2!−6!×3!=14400

$${METHOD}\:{I} \\ $$$${between}\:{two}\:{boys}\:{there}\:{must}\:{be}\:{at}\:{least} \\ $$$${one}\:{girl}. \\ $$$${we}\:{arrange}\:{at}\:{first}\:{the}\:\mathrm{5}\:{girls},\:{there} \\ $$$${are}\:\mathrm{5}!\:{ways}. \\ $$$$\Box{G}\Box{G}\Box{G}\Box{G}\Box{G}\Box \\ $$$${the}\:\mathrm{3}\:{boys}\:{can}\:{now}\:{take}\:{three}\:{of} \\ $$$$\mathrm{6}\:{positions}\:\left(\Box\right).\:{there}\:{are}\:{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{3}!\:{ways}. \\ $$$${so}\:{totally}\:{there}\:{are}\:{C}_{\mathrm{3}} ^{\mathrm{6}} ×\mathrm{3}!×\mathrm{5}!=\mathrm{14400} \\ $$$${ways}. \\ $$$$ \\ $$$${METHOD}\:{II} \\ $$$${to}\:{arrange}\:\mathrm{8}\:{children}\:{there}\:{are}\:\mathrm{8}! \\ $$$${ways}. \\ $$$$ \\ $$$${to}\:{arrange}\:{them}\:{such}\:{that}\:{three}\:{boys} \\ $$$${are}\:{together},\:{there}\:{are}\:\mathrm{6}!×\mathrm{3}!\:{ways}. \\ $$$$ \\ $$$${to}\:{arrange}\:{them}\:{such}\:{that}\:{two}\:{and} \\ $$$${only}\:{two}\:{boys}\:{are}\:{next}\:{to}\:{each}\:{other}, \\ $$$${there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{3}} ×\left(\mathrm{7}!−\mathrm{2}×\mathrm{6}!\right)×\mathrm{2}!\:{ways}. \\ $$$$ \\ $$$${total}\:{valid}\:{arrangements}: \\ $$$$\mathrm{8}!−{C}_{\mathrm{2}} ^{\mathrm{3}} ×\left(\mathrm{7}!−\mathrm{2}×\mathrm{6}!\right)×\mathrm{2}!−\mathrm{6}!×\mathrm{3}!=\mathrm{14400} \\ $$

Commented by john santu last updated on 28/May/20

oo yes...your right. i′m using method II but forgot something. ok thanks you

$$\mathrm{oo}\:\mathrm{yes}…\mathrm{your}\:\mathrm{right}.\:\mathrm{i}'\mathrm{m}\:\mathrm{using}\:\mathrm{method}\:\mathrm{II} \\ $$$$\mathrm{but}\:\mathrm{forgot}\:\mathrm{something}.\:\mathrm{ok}\:\mathrm{thanks}\:\mathrm{you} \\ $$

Commented by bobhans last updated on 28/May/20

how with ■GG■G■GG with ■GGG■G■G ? your answer included it ?

$$\mathrm{how}\:\mathrm{with}\:\blacksquare\mathrm{GG}\blacksquare\mathrm{G}\blacksquare\mathrm{GG} \\ $$$$\mathrm{with}\:\blacksquare\mathrm{GGG}\blacksquare\mathrm{G}\blacksquare\mathrm{G}\:?\: \\ $$$$\mathrm{your}\:\mathrm{answer}\:\mathrm{included}\:\mathrm{it}\:? \\ $$

Commented by mr W last updated on 28/May/20

yes, included. the boys can take any three positions from the 6 □ in: □G□G□G□G□G□ examples: ■G□G■G■G□G□ ■G□G□G■G■G□

$${yes},\:{included}. \\ $$$${the}\:{boys}\:{can}\:{take}\:{any}\:{three}\:{positions} \\ $$$${from}\:{the}\:\mathrm{6}\:\Box\:{in}: \\ $$$$\Box{G}\Box{G}\Box{G}\Box{G}\Box{G}\Box \\ $$$${examples}: \\ $$$$\blacksquare{G}\Box{G}\blacksquare{G}\blacksquare{G}\Box{G}\Box \\ $$$$\blacksquare{G}\Box{G}\Box{G}\blacksquare{G}\blacksquare{G}\Box \\ $$

Commented by i jagooll last updated on 28/May/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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