The number of numbers of 9 different 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

2 (4!)

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(4!)^{2}

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8!

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none of these

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Solution

The correct option is B $(4!)^{2}$
So as per the question the only number that can be in the middle is $5$ .
Thus $1, 2, 3, 4$ can be permuted in $1 s t$ $4$ boxes and

$6, 7, 8, 9$ can be permuted in last $4$ boxes
Number. of such numbers= $4! \times 4! = {4!}^{2}$
Hence, option 'B' is correct.

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The number of numbers of 9 different 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

The correct option is B (4!)2So as per the question the only number that can be in the middle is 5.Thus 1,2,3,4 can be permuted in 1st 4 boxes and 6,7,8,9 can be permuted in last 4 boxesNumber. of such numbers=4!×4!=4!2Hence, option 'B' is correct.

The correct option is B $(4!)^{2}$
So as per the question the only number that can be in the middle is $5$ .
Thus $1, 2, 3, 4$ can be permuted in $1 s t$ $4$ boxes and

$6, 7, 8, 9$ can be permuted in last $4$ boxes
Number. of such numbers= $4! \times 4! = {4!}^{2}$
Hence, option 'B' is correct.