Question

The number of nine 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

• A. $2(4!)$
• B. $\frac{3(7!)}{2}$
• C. $2(7!)$
• D. ${}^4{P}_4{\times }^4{P}_4$

Answer 1

The correct option is D ${}^4{P}_4{\times }^4{P}_4$

According to given conditions, number can be formed by the following format:
The required number of numbers is ${}^4{P}_4{\times }^4{P}_4$.