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×(4!)2

Answer 1

In this problem we will use permutation and combination concepts to allocate the digits at different places keeping in mind the condition stated in the problem

Complete step by step solution:
Given: The number of different nine-digit numbers formed by using the digits 1 to 9 without repetition. 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

Firstly, we will find the possibility of the middle number in this question
Since, 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.

Therefore, middle digit has to be the middle number among numbers from 1 to 9 (1,2,3,4,5,6,7,8,9)

That is, the middle digit should be 5 to satisfy the condition.
Now, number of possible ways to fill first four digits $\; = {\rm{ }}4!$ (Since only 1,2,3,4 can be filled in first four places as all the digits in the first four places are less than the digit in the middle)

Number of possible ways to fill last four digits $\; = {\rm{ }}4!$ (Since only 6,7,8,9 can be filled in last four places as all the digits in the last four places are greater than the digit in the middle)

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

Hence, option (B) is the correct answer.

Note: In such kind of problem, we find the number of possible ways the numbers can be arranged according to the condition provided in the question.