Question

Ten persons, amongst whom are A, B and C to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is

• A. $\frac{10!}{6}$
• B. 3! 7!
• C. ¹⁰P₃
• D. None of these

Answer 1

Answer: (A)

$\frac{10!}{6}$

Answer 2

For A, B C to speak in order of alphabets, 3 places out of 10 may be chosen first in ¹⁰C₃ ways.

The remaining 7 persons can speak in 7! Ways. Hence, the number of ways in which all the 10 person can speak is ¹⁰C₃. 7! = $\frac{10!}{3}$= $\frac{10!}{6}$