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Q.

 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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a

2(7!)

b

 4P4×4P4

c

3(7!)/2

d

2(4!)

answer is D.

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Detailed Solution

According to given conditions, numbers can be formed by the following format:

Question Image









The required number of numbers is  4P4×4P4. 

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