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

The total number of ways to arrange 6 distinct people in a row is 6 factorial, which equals 720.

The problem imposes a specific relative order between three people: A, B, and C. The condition states that B cannot be ahead of A and C cannot be ahead of B. This means their relative positions must strictly be in the order A followed by B followed by C.

In any random arrangement of 6 people, there are 3 factorial (which is 6) possible ways A, B, and C can be positioned relative to each other. These ways are:

1. A-B-C
2. A-C-B
3. B-A-C
4. B-C-A
5. C-A-B
6. C-B-A

Out of these 6 possible relative orders, only 1 satisfies the condition (A-B-C). Since every relative order is equally likely across all 720 permutations, we simply divide the total arrangements by the number of possible relative orders of the three restricted individuals.

Calculation: 720 divided by 6 equals 120.

Therefore, there are 120 ways to arrange the 6 people such that the relative order A-B-C is maintained. Correct answer is C.