Question of the Week- 25 (For a meeting, 9 speakers must be .........)

For a meeting, 9 speakers must be selected to give a speech, one after the other, from a group of 12 people. What is the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C?

Firstly, 9 speakers are selected from 12. This can be done in 12*11*10 ways or \(^{12}P_9\) ways.
Now A,B and C are in the group of 9 people. and B should speak before A and C. This can be done in 7 ways.

A and C be seated after B in the following ways:

B_ _ _ _ _ _ _ _ = \(^8C_2*2!\) (Selecting 2 places among 8 and arranging A and C within their places)
_B_ _ _ _ _ _ _ = \(^7C_2*2!\)
_ _B_ _ _ _ _ _ = \(^6C_2*2!\)
_ _ _B_ _ _ _ _ = \(^5C_2*2!\)
_ _ _ _B_ _ _ _ = \(^4C_2*2!\)
_ _ _ _ _B_ _ _ = \(^3C_2*2!\)
_ _ _ _ _ _B_ _ = \(^2C_2*2!\)

Now B can't be in 8th place as it leaves only one place after B in which both A and C can't sit.

\(\frac{^8C_2*2! + ^7C_2*2! + ^6C_2*2! + ^5C_2*2! + ^4C_2*2! + ^3C_2*2! + ^2C_2*2!}{12*11*10}\)

\(\frac{(7*8)+(6*7)+(5*6)+(4*5)+(3*4)+(2*3)+(1*2)}{12*11*10}\)

\(\frac{56+42+30+20+12+6+2}{12*11*10}\)

\(\frac{168}{12*11*10}\)

\(\frac{7}{11*5}\)

Though it may seem lengthy, if you are confident about the procedure then it should not take more than 1 min 30 seconds to solve this.

OPTION: E