Question

$A,B,C$ are three persons among $7$ persons who speak at a function. The number of ways in which it can be done. If '$A$' speaks before '$B$' and '$B$' speaks before '$C$' is

• A. $820$
• B. $830$
• C. $840$
• D. $850$

Answer 1

Answer: (C)

$840$

It is given that $A$ comes before $B$ and $B$ comes before $C$.

Therefore, $A$ comes before $C$.

And it is not necessary for the $3$ to be together, i.e., speak exactly after one another.

The required permutation will be ${}^7{P}_{7-3}$ $={}^7{P}_4$

$=\frac{7!}{3!}$

$=4\left(5\right)\left(6\right)\left(7\right)$

$=7.\left(120\right)$

$=840$