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An unlimited number of coupons bearing the letters A, B and C are available. What is the possible number of ways of choosing 3 of these coupons so that they cannot be used to spell BAC?


Options:
A .   15
B .   18
C .   21
D .   27
E .   23
Answer: Option C
:
C

Each of the 3 places can take 3 letters 27. But we don't want the combination (A, B, C) 3! = 6 are out 27-6 = 21. Hence, choice (c) is the right answer.



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