The number of nine nonzero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is

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2(4!)

3(7!)/2

2(7!)

4P4×4P4

answer is D.

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Detailed Solution

According to given conditions, numbers can be formed by the following format:The required number of numbers is 4P4×4P4.

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