Question

Heloo friends

six persons A, B, C, D ,E and F are to be seated at a circular table . 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 ?

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Answer 1

$\bold \orange{hello \: dosto !!}$
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$\bold \green{solution} \:$
:- When A has B or C to his right we have AB or AC .
When B has C or D to jis right We have BC or BD
thus => we must have ABC or ABD or AC and BD

for ABCD, E , F on a circle number of ways = 3! = 6

for ABDC , E, F on a circle number of ways = 3! = 6

for AC , BDE,F the number of ways 3! = 6

=> total = 18

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$\boxed{hope \: it \: helps \: you}$

$\pink{@Rajukumar111}$

Answer 2

Answer:

Say B on right side of A

D or C will be right of B

This can be done in 2! ways .

Remaining three places can be filled in 3! ways

=2!× 3! = 12 ways

Say C on right of A

B can be filled in 3 places ( B next should be D)

B immediately next is D

A,B ,C,D are filled in 3 ways

Remaining two places in 2 ways

Total= 6 ways

Total number = 12 +6= 18 ways