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.