First slide

Combinations

Question

Among 10 persons, A, B, C are to speak at a function. The number of ways in which it can be don e if A wants to speak before B and B wants to speak before C is

Moderate

A

10!/24
B

9!/6
C

10!/6
D

none of these

Solution

Places for A, B, C can be chosen in ¹⁰C₃ ways. Remaining 7 persons can speak in 7! ways. Hence, the number of ways in which they can speak is ¹⁰C₃ $\times$ 7! = 10!/6.

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