Question

The digits 1,2,3,4,5,6,7,8,9 are written in random order to form a nine digit number. Find the probability that this number is divisible by 4:

• A. $\frac{4}{9}$
• B. $\frac{2}{9}$
• C. $\frac{17}{81}$
• D. $noneofthese$

Answer 1

The correct option is B $\frac{2}{9}$

Total possible nine digit numbers = 9!
Out of these 9! numbers only those numbers are divisible 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 possible numbers of last two digits are 12,32,52,72,92,24,64,84,16,36,56,76,96,28,48,68.
Thus 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 digits numbers divisible by 4 is 16 $\times$ 7!.
Hence, required probability = $\frac{16\times 7!}{9!}=\frac{2}{9}$