Question

Six persons A, B , C, D, E and F are to be seated at a circular table . Find the number of ways this can be done if A must have either B or C on his right and B must have either C or D on his right.
a 3. b 6. c 12. d 18​

Answer 1

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$\Large{\fcolorbox{green}{grey}{ \bf{\underline{\green{ \red{Correct }Answer\pink ✎}}}}}$

$\LARGE\bold\red{ is:}$

Option (d) 18 ✔

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$\fcolorbox{green}{red}{\Large{\fcolorbox{cyan}{grey}{\bf{\underline{\green{ \red{EX}\orange{PL}AN\purple{AT}\pink{IO}\red{N} \orange↝}}}}}}$

❀There are 3 possible cases ☟

$\large{\fcolorbox{pink}{grey}{\red{\underline{{\pink{Case \red 1.}}}}}}\Large :$ when A, B and C are sitting together (in the same order)

⇨Possible combinations = (4-1)!=6

$\large{\fcolorbox{pink}{grey}{\red{\underline{{\pink{Case \red 2.}}}}}}\Large :$ when A, B and D are sitting together (in the same order)

⇨Possible combinations= (4-1)!=6

$\large{\fcolorbox{pink}{grey}{\red{\underline{{\pink{Case \red 3.}}}}}}\Large :$ when A and C are sitting together and B and D are sitting together

⇨Possible combinations= (4-1)!=6

$\Large\mathtt{\green{\underline{\rm Henc\rm E}}\pink ࿐}$

The number of ways this can be done if A must have either B or C on his right and B must have either C or D on his right is: 6+6+6=18

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❖ ─────$\Large{\mathfrak{\red{\underline{ \blue{\underline{\green{\underline{\purple E \orange N \pink D}}}}}}}}$───── ❖

Answer 2

$option \: d \: is \: the \: correc \: answer$