Three dice are rolled. The number of The number of ways in which ten candidates A1 , A2 , ...., A10 can be ranked so that A1 is always above
The number of ways in which ten candidates A1 , A2 , ...., A10 can be ranked so that A1 is always above A2 is
(a) 10!/2 (b) 10 ! (c) 9 ! (d) 8!/2
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(a) Ten candidates can be ranked in 10! ways. In half of these ways A1 is above A2 and in another half A2 is
above A1. So, required number of ways is 10!/2 .
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