There are 6 persons-A, B, C, D, E and F. They are to be seated in a row such that B never sits any where ahead of A and C never sits any where ahead of B. In how many different ways can this be done?

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Submitted by examrobotsa on

Q: 82 (IAS/2005)
There are 6 persons-A, B, C, D, E and F. They are to be seated in a row such that B never sits any where ahead of A and C never sits any where ahead of B. In how many different ways can this be done?

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,3,10,2,7,3,1

keywords: 

{'row': [0, 0, 1, 0], 'many different ways': [0, 0, 8, 0], 'none': [2, 0, 8, 15], 'persons': [4, 4, 9, 10]}

The rule implies that C must always sit behind B, and B must always be behind A. This placement constraint leaves their order as A, B and C. So, they can be thought of as a single unit. Now, including D, E, and F, we have a total of 4 units to be seated. These can be arranged in 4! ways which equals 24. Within the ABC unit, A, B, and C can shuffle between themselves in 3! ways which equals 6. By the multiplication principle, the total ways equals 24*6=144.

The options provided are 60, 72, 120, and None of the Above. But none of the options match the calculated 144 ways.

Alert - correct answer should be "None of the Above". The numerical answer being 144 is absent in the list.