1. Tardigrade
  2. Question
  3. Mathematics
  4. 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

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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

Permutations and Combinations

Solution:

For $A , B , C$ to speak in order of alphabets, $3$ places out of $10$ may be chosen first in ${ }^{10} C _{3}$ ways.
The remaining $7$ persons can speak in $7 !$ ways.
Hence, the number of ways in which all the $10$
persons can speak is ${ }^{10} C _{3} \cdot 7 !$
$=\frac{10 !}{3 !}=\frac{10 !}{6}$.