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 ${}^{\prime }{A}^{\prime }$ speaks before ${}^{\prime }{B}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ speaks before ${}^{\prime }{C}^{\prime }$ is

Answer 1

Number of ways in which $7$ persons can speak in $7$ places such that order of three particular persons $A,B,C$ is fixed = $\frac{7!}{3!}=840$.

ALTERNATE SOLUTION:
Selection $3$ places from $7$ places can be done ${=}^7{C}_3$ways.
No of ways in which three particular persons $A,B,C$ can speak in this function ${=}^7{C}_3$(as order is fixed)
Now, remaining $4$ people can speak in $4$ places in $4!$ ways
$\therefore$ Required number of ways ${=}^7{C}_3\times 4!=840$