Four speakers will address a meeting where speaker $Q$ will always speak after $P$ . Then, the number of ways in which the order of speakers can be prepared is

256

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128

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24

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12

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Solution

The correct option is C $12$
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 $Q$
$∴$ Required number of ways $= \frac{24}{2} = 12$

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Four speakers will address a meeting where speaker Q will always speak after P. Then, the number of ways in which the order of speakers can be prepared is

The correct option is C 12Four 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∴ Required number of ways =242=12

Four speakers will address a meeting where speaker $Q$ will always speak after $P$ . Then, the number of ways in which the order of speakers can be prepared is

The correct option is C $12$
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 $Q$
$∴$ Required number of ways $= \frac{24}{2} = 12$