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 the digit in the middle is

Question

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 the digit in the middle is (A) 2(4 !) (B) (4 !)2 (C) 8 ! (D) None of these

Tardigrade Admin · Accepted Answer

x 1 x 2 x 3 x 4 underlinex5 x 6 x 7 x 8 x 9. Under the given situation x 5 can be 5 only. The selection for x 1, x 2, x 3, x 4 must be from 1,2,3,4, so they can be arranged 4 ! ways. Again the selection of x6, x7, x8, x9 must be from 6,7,8,9 so they can be arranged in 4 ! ways. Desired number of ways =(4 !)(4 !)=(4 !)2

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 the digit in the middle is

$2 (4!)$

$(4!)^{2}$

$8!$

None of these

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 the digit in the middle is

2(4!)

(4!)2

8!

None of these

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 the digit in the middle is

$2 (4!)$

$(4!)^{2}$

$8!$