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Question

Number of permutations 1,2,3,4,5,6,7,8 and 9 taken all at a time are such that the digit.
1 appearing somewhere to the left of 2,
3 appearing to the left of 4 and
5 somewhere to the left of 6, is?
(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!
There are exactly 9! ways in which we can arrange this set out of which 1 is to left of 2 in half of them out of which exactly half of them has 3 to left of 4 and from that exactly half of them has 5 to the left of 6. Hence,
answer = (12)3.9!=9.7!

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