Question

Three persons A,B and C are to speak at a function along with five others. If they all speak in random order, the probability that A speaks before B and B speaks before C is

Answer 1

Hint: First we find the total arrangement of the speakers that can speak and then we find the probability of the rest of the other persons aside from A, B and C and then we use the formula as:
\[\dfrac{{}^{\text{Total persons}}{{C}_{\text{A,B and C}}}\times \left( \text{OP} \right)!}{{}^{\text{Total persons}}{{P}_{\text{Total persons}}}}\]
where OP is the total number of people other than A, B and C.

Complete step by step answer:
According to the question given, the total number of people to speak at the function is given as eight persons. Now to see the permutation of speaking of three persons i.e. A, B, C and others in random order is given as:
\[\Rightarrow {}^{8}{{P}_{8}}=8!\]
The total number of ways in which the three persons can form probability with the total number of people being A, B, C and five others is \[^{8}{{C}_{3}}\].
The total number of probability in which the other five people can speak is \[5!\]
Hence, using the number of probability we can say that with A, B, C and other five people, A speaks before B and B speaks before C is equal to:
\[\Rightarrow \dfrac{{}^{8}{{C}_{3}}\times 5!}{{}^{8}{{P}_{8}}}\]
\[\Rightarrow \dfrac{{}^{8}{{C}_{3}}\times 5!}{8!}\]
Therefore, the total number of ways or probability in which 8 people can speak with A, B and C first is \[\dfrac{1}{6}\].

Note: The first part that is for 8 person can speaking at a certain arrangement is \[{}^{8}{{P}_{8}}=8!\]; meaning we use permutation and not combination as we need the total ways of all the speaker can speak, we do not use combination as combination is used when we select from all the possibilities that comes in the second part where we need the answer.