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Bunuel
Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

A, 6
B. 12
C. 18
D. 24
E. 30


ABCD
CD=X; 2! WAYS
XAB ; 3! WAYS
3!*2! ; 12
IMO B
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Bunuel
Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

A, 6
B. 12
C. 18
D. 24
E. 30

Number of ways of not sitting together = Total ways - When both sit together

Total ways = 4! = 24 ways

When C&D are together—> CD, A, B (consider C&D as one unit)
Number of ways = 3!*2! = 6*2 = 12 ways

Not together = 24 - 12 = 12 ways

Option B.

Pls Hit kudos if you like the solution

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Bunuel
Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

A, 6
B. 12
C. 18
D. 24
E. 30

No of ways to arrange people = 4! = 24 ways.

Bunuel.. Can we consider the case that half of the time CD will sit together and the other half they won't? So then the answer will be 12?
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