267 The number of ways in which 12 dice can be thrown so that each face occur exactly twice is , (a) 6 (b) 2 (c) 6' 12!(d) None

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The sample space will be ( (6)^{wedge} 12 ). The difficult part is the favourable cases part. Let's see it this way: All the faces come Twice ie the events contain ( 1 mathrm{s}(2 text { times }), 2 mathrm{s}(2 text { times }), 3 mathrm{s}(2 text { times }), 4 mathrm{s}(2 text { times }), 5 mathrm{s}(2 text { times }) ) and ( 6 mathrm{s}(2 text { times }) ) So the cases will be permutations of {4,5,4,1,2,2,5,6,1,3,3,6} So number of permutations of 6 objects each repeating twice making 12 objects.... is ( 12 ! /(2 !)^{wedge} 6=7484400 ) Required probability ( =7484400 / 6^{wedge} 12 )

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