The digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 are written in random order to form a nine digits number. Let p be the probability that this number is divisible by 36, then 9p is?

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answer is C.

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Detailed Solution

We know from the definition of probability that if there is

n (A)

number of ways of event A occurring (or number of favourable outcomes) and

n (S)

is the size of the sample space (number of all possible outcomes) then the probability of the event A occurring is given by

P (A) = n (A) n (S)

We are given in the question that the digits 1, 2,3,4,5,6,7,8, and 9 are written in random order to form a nine-digit number. So the number of 9-digit numbers from the digits 1, 2,3,4,5,6,7,8, and 9 is the number of ways we can arrange 9 distinct digits is 9!. So the number of all possible outcomes is

n (S) = 9!

Let us denote the event of getting a number divisible by 36 as AA. We know that a number that is divisible by 36 has to be divisible by relative primes 9 and 4. Since of sum of the digits 1+2+3...+9=36 is divisible by 9, all 9-digits number formed will divisible by 9. So we only need to find a number of 9-digit numbers which are divisible by 4.
We know that a number is divisible by 4 when the number formed by the last two digits is divisible. So the possible numbers than can be formed last two digits are