Question -6

Let the nine digit number be

${ a }_{ 1 }{ a }_{ 2 }{ a }_{ 3 }{ a }_{ 4 }{ a }_{ 5 }{ a }_{ 6 }{ a }_{ 7 }{ a }_{ 8 }{ a }_{ 9 }$

The total number of numbers that can be formed = $9 \text{!} =\boxed {362880}$

For a number to be divisible by $4$ , the last two digits need to be divisible by $4$

Thus the last two digits, ${ a }_{ 8 }{ a }_{ 9 }$ can have the following arrangements-

$12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96 = 16$

We cannot have $04, 08, 20, 40, 60, 80$ since we do not have $0$

We cannot have $44, 88$ since repetition is not allowed.

After choosing ${ a }_{ 8 }{ a }_{ 9 }$ , we have seven spaces left which can be arranged in $7 \text{!}$ ways which can be taken $16$ times hence

$7 \text{!}\times 16 = \boxed {80640}$

Thus the Probability is $\frac {80640}{362880} = \boxed {\frac {2}{9}}$

Firstly we find the number of way to form ${even}$ number using $1,2,3,........,9$ without repetition : To form even number last digit must be $2,4,6 \ or \ 8$ . That mean there are $4$ choices. So number of way to form even number is $4\times 8!$ . Therefor the number of way to form numbers divisible by $4$ is ,

$\frac{1}{2} \times number \ of \ even \ numbers$ = $2\times 8!$

So the required probability is , $\frac{2\times 8!}{9!} = \frac{2}{9}$

Python:

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from itertools import permutations
from fractions import Fraction as frac
digits = '123456789'
count = 0
total = 0
for combo in permutations(digits):
    total += 1
    test = ''.join(combo)
    if len(test) == 9:
        if int(test) % 4 == 0:
            count += 1
print frac(count,total)