Six speakers are to address a seminar one after another. Three of them — the chief guest, the convener and the department head — must speak in that relative order, the chief guest before the convener and the convener before the department head, though any of the other speakers may take the floor in between. In how many different orders can the six speakers be scheduled?

Schedule all six freely: 6! = 720 orders. In any one of them the three named speakers fall in some order among themselves, and by symmetry each of the 3! = 6 possible internal orders occurs equally often. Only ONE of those six is the required order, so the answer is 720/3! = 120. Trap 720 forgets the condition. Trap 24 = 4! treats the three as a single block, which wrongly forces them to speak consecutively — the question expressly allows other speakers in between. Trap 144 = 720/5 divides by the wrong number; the divisor must be the count of internal orders, 3! = 6.