Correct Answer - C
There are
.
9
P
9
=
9
!
.
ways of arranging the given digits to form a nine digit number. So, total number of 9 digit numbers formed by the given digits =9!.
Out of these 9! Numbers only those numbers are divisble by 4 which have their last digits as even natural number and the numbers formed by their last two digits are divisible by 4.
The various possibilities of last two digits are :
12, 32, 52, 72, 92
24, 64, 84
16, 36, 56, 76, 96
28, 48, 68
This means that there are 16 ways of choosing the last two digits. Corresponding to each of these ways the remaining 7 digits can be arranged in 7! ways. Therefore, the total number of 9 digit numbers divisble by 4 is
16
×
7
!
16
Hence, required probability
=
16
×
7
!
9
!
=
2
9
=