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Question

Number of permutations of 1,2,3,4,5,6,7,8 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 B 9.7!
There are 91 permutations possible of the set [1,2,3,4,5,6,7,8,9}
1 is to the left of 2 in exactly half of these.
3 is to the left of 4 in exactly half of the permutations and similarly half of the permutations contain 5 to the left of 6. There by the total permutations for the above three independent cases is 12.12.12.91=18.9.8.7!=9×7!
Alternatively,
Consider the arrangement {1,2}, we can choose 9C2 places for such an arrangement. Then 7C2 places for {3,4} and then 5C2 places for {s,b} and then in 3! ways we can arrange 7,8 & 9 giving used total of
9C2×7C2×5C2×3!=9×7!
Note : {1,2} does not mean that 1,2 are together. It just implies that 1 is to the left of 2.

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