Four speakers will address a meeting where speaker Q will always speak after speaker P. Then the number of ways in which the order of speakers can be prepared is:
A. 256
B. 128
C. 24
D. 12

Hint: To solve this question, we will first assume the other two speakers to be R and S. Then, we will take all the cases in which P is at the first, second, third, and fourth position, and with respect to these, Q speaks after it. Then we will arrange Q, R, and S according to the given condition and then add the outcomes of all these cases. Hence, we will get our answer.

Complete step-by-step solution
We here need to find the number of ways Q will always speak after P among the four speakers. Now, for this let us first assume the other speakers to be R and S.
Now, P can be any of the first, second, third, or fourth speakers and Q will always speak after P.
So let us take all these cases separately.
Case-I: P at position-1
Now, since P is at position 1, Q will automatically be after it and hence can take any of the next three positions. R and S will also have the same choices as that Q. Thus, these three speakers, Q, R, and S can speak in any order. So, we have to arrange these 3 different speakers in 3 positions.
Now, we know that when n distinct things have to be arranged in n different ways, it is given by the formula n!.
Thus, these three speakers can speak in the following ways:
$\begin{align}
  & 3! \\
 & \Rightarrow 6 \\
\end{align}$
Case-II: P at position-2
Now, since P is at position 2, Q can take the third or fourth position only since it has to speak after P. Thus, Q has two choices. Simultaneously with this, R and S have been left with two choices.
Thus, in this case, the rest three speakers can speak in the following ways:
$\begin{align}
  & 2\times 2! \\
 & \Rightarrow 2\times 2 \\
 & \Rightarrow 4 \\
\end{align}$
Case-III: P at position-3
Now, since P is in the third position, Q can only speak at the last. Thus it has only one choice. Simultaneously with this, R and S have the first two positions to speak in.
Thus, in this case, the rest three speakers can speak in the following ways:
\[\begin{align}
  & 1\times 2! \\
 & \Rightarrow 1\times 2 \\
 & \Rightarrow 2 \\
\end{align}\]
Case-IV: P at position-4
Now, since P at the last position, Q will always speak after P. Hence, there wouldn’t be any way in this case to meet the required possibility.
Now, if we add the outcomes of all these cases, we will get the required number of ways in which P will always speak before Q.
Thus, we get:
$\begin{align}
  & 6+4+2+0 \\
 & \therefore 12 \\
\end{align}$
Thus, option (C) is the correct option.

Note: We have here calculated the answer by the complete process but we can solve this question by making use of the options given in competitive exams to save time. This can be done as follows:
Now, there are four speakers so the maximum ways in which they can speak is given as:
$\begin{align}
  & 4! \\
 & \Rightarrow 24 \\
\end{align}$
Thus, the first two options are eliminated as they are more than 24.
Now, among these 24 ways, there will be cases in which P will speak before Q and also their vice versa. Thus, the required answer is less than 24 which is only option D, i.e. 12.
Hence, the option (D) is the correct option.