WBJEE · Maths · Permutation Combination
Four speakers will address a meeting where speaker \(Q\) will always speak before \(P .\) Then, the number of ways in which the order of speakers can be prepared is
- A 256
- B 128
- C 24
- D 12
Answer & Solution
Correct Answer
(D) 12
Step-by-step Solution
Detailed explanation
Four speakers will address the meeting in \(4 !\) ways \(=24\) different ways in which half number of cases will be such that \(P\) speaks before \(Q\) and half number of case will be such that \(P\) speaks after \(Q\) \(\therefore\) Required number of ways \(=\frac{24}{2}=12\)
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