Question

The number of different nine digit numbers formed by using the digits 1 to 9 without repetition, 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. $2\times (4!{)}^2$

Answer 1

The correct option is B $(4!{)}^2$

The given condition will be satisfied if the digit $5$ occupies the middle position as shown.

The first four places should be occupied by the digits $1,2,3,4$. The digits can be arranged in $4!$ ways.

Similarly, the digits at the last four places $6,7,8,9$ can be arranged among themselves in $4!$ ways.

Hence, the total number of nine digit numbers$=4!\times 4!=(4!{)}^2$