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Question

Number of permutations of 1,2,3,4,5,6,7,8,9 taken all at a time are such that the digit
1 appearing some where to the left of 2
3 appearing to the left of 4
5 some where to the left of 6
(e.g.815723946 would be one such permutation)

A
9.7!
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B
8!
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C
5!.4!
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D
8!-4!
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Solution

The correct option is A 9.7!
Total no. of permutations =9!
In these, half have 1 in left of 2,
half have 3 in left of 4,
half have 5 in left of 6,
No. of required permutations =12×12×12×9!
=9×7!
Hence, the answer is 9×7!.

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