The number of numbers with 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 the digit in the middle is

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144

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288

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576

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Solution

The correct option is D 576
The middle digit must be $5, 5^{t h}$ place, first four places can be filled with 1,2,3,4 and last four places can be filled with 6,7,8,9. Hence required number of number of 9 digits $= 4! 4! = (4!)^{2} = (24)^{2} = 576$

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♠

\to

♠

is a

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-digit number.

\to

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and their sum is

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The number of numbers with 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 the digit in the middle is

The correct option is D 576The middle digit must be 5,5th place, first four places can be filled with 1,2,3,4 and last four places can be filled with 6,7,8,9. Hence required number of number of 9 digits =4!4!=(4!)2=(24)2=576

The correct option is D 576
The middle digit must be $5, 5^{t h}$ place, first four places can be filled with 1,2,3,4 and last four places can be filled with 6,7,8,9. Hence required number of number of 9 digits $= 4! 4! = (4!)^{2} = (24)^{2} = 576$