wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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!)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(4!)2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
8!
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2×(4!)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

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!×4!=(4!)2


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Permutations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon