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Q.

The number of numbers of 9 different non zero 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 (4!)

b

8!

c

2 (8!)

d

{(4!)}^{2}

answer is B.

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Detailed Solution

Given digits

1, 2, 3, \dots, 9

Question Image

Required number of ways $=^{4} P_{4} .^{4} P_{4} = {(4!)}^{2}$

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The number of numbers of 9 different non zero 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