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(a) $\dfrac{53\times 4!}{5\times 6!} $

(b) $\dfrac{53}{180} $

(c) $\dfrac{53!}{6\times 6!} $

(d) None of these

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We should know that two or more than two digit numbers are divisible by 4 if it’s last two digits of the number are divisible by 4,

Digits that are given to us are 0,1,2,3,4,8,9 so possible last two digits can be,

04, 08, 12, 20, 24, 28, 32, 40, 48, 80, 84, 92.

Case 1: we will suppose that 0 is in last two digits of the number, then:

Different cases that will be possible are 04, 08, 20, 40, 80, hence we get

Total number of different cases will be 5 when 0 included in last two digits, and

We know that, first five digits can be arranged in 5! Ways, hence, we get

Total number of ways to form a number according to case 1 will be = 5 $\times $ 5!

Case 2: we will suppose when 0 in neither at last two digits and nor at first place then,

There are four places where 0 can be placed i.e. are from ${{2}^{nd}} $ to ${{5}^{th}} $ so, we get

Total number of ways zero can be placed will be = 4

And we can conclude that,

last two digits can be placed in 7 ways = 12, 24, 28, 32, 48, 84, 92

Out of the first five digits four digits can be arranged in 4! Ways (because 0 is excluded), and we get

Total number of cases according to case 2 will be = 4 $\times $ 4! $\times $ 7

And also we will find out total number of 7 digit numbers that can be formed using given digits will be,

= 7! – 6! = 6 $\times $ 6!

As 6! Is subtracted from 7! As we considered a case when 0 can’t be placed at first placed and in that case number of cases will be 6! otherwise 7! cases are there for 7 numbers.

Now, we know

Probability = $\dfrac{favourable\,cases}{total\,number\,of\,cases}=\dfrac{case1+case2}{total\,number\,of\,cases} $

= $\dfrac{(5\times 5!)+(4\times 7\times 4!)}{6\times 6!} $

After further solving the above expression, we get

Probability = $\dfrac{53}{180} $