(D) The total number of ways to arrange $4$ speakers is $4! = 4 \times 3 \times 2 \times 1 = 24$.In any arrangement, there are only two possibilities

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(D) The total number of ways to arrange $4$ speakers is $4! = 4 \times 3 \times 2 \times 1 = 24$.In any arrangement, there are only two possibilities for the relative order of speakers $P$ and $Q$: either $P$ speaks before $Q$ or $Q$ speaks before $P$.Since these two cases are equally likely, exactly half of the total arrangements will have $Q$ speaking before $P$.Therefore, the required number of ways is $\frac{24}{2} = 12$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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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

WBJEE 2012 All WBJEE

A
$256$
B
$128$
C
$24$
D
$12$

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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