Multiple choice

In a function, 10 persons including A, B and C have to perform. If A has to perform before B and B has to perform before C, how many different ways are there to arrange the order of their performances?

  1. 10 ! 6

  2. 10C3

  3. 10P3 × 7!

  4. 10! × 3!

  5. 3! × 7!
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A Correct answer
Explanation

Total arrangements of 10 people is 10!. In any arrangement, the relative order of A, B, and C can be in 3! = 6 ways. Only one of these ways (A before B before C) is valid. Thus, the number of valid ways is 10! / 3! = 10 * 9 * 8 * 7! / 6 = 10 * 9 * 8 * 7! / 6 = 120 * 7! = 10! / 6.

AI explanation

The total number of ways to arrange the 10 persons without any conditions is 10!. For any specific positions chosen for A, B, and C among the 10 slots, there is only 1 valid sequence where A precedes B and B precedes C, out of 3! = 6 total possible sequences for the trio. We divide the total unrestricted arrangements by 6 to account for this condition, yielding 10!/6.

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