Question

The probability that a \[9\] digit number formed by $1,2,3,...,9$ without repetition is divisible by $36$ is?
A) $\dfrac{1}{7}$
B) $\dfrac{2}{9}$
C) $\dfrac{1}{3}$
D) $\dfrac{4}{9}$

Answer 1

Hint: Nine digit numbers can be formed using nine digits without repetition in $9!$ ways. Factorising $36$ as $9 \times 4$ we can find how many among them are multiples of $36$. Then we can see the probability using a favourable number and total number.

Useful formula:
For some value $n$, $n$ objects can be filled in $n$ positions in $n!$ ways.
Probability is obtained by dividing favourable number by total number.

Complete step-by-step answer:
We know $n$ objects can be filled in $n$ positions in $n!$ ways.
We can form a nine digit number using the digits $1,2,3,...,9$ without repetition in $9!$ ways.
Now we have to find out how many among them are divisible by $36$.
We know, $36 = 9 \times 4$
So, for a number to be a multiple of $36$, it must be multiple of $9$ as well as $4$.
We know that for a number to be a multiple of $9$, the sum of the digits of the number must be multiple of $9$.
But we have, $1 + 2 + 3 + ... + 9 = 45$ and $45$ is a multiple of $9$.
So any nine digit number formed using $1,2,3,...,9$ without repetition will be a multiple of nine.
Now for a number to be a multiple of four, the last two digits must be a multiple of four.
So possibilities of the last two digits are $12,16,24,28,32,36,48,52,56,64,68,72,76,84,92,96$.
In total there are $16$ possibilities.
At the same time the first seven digits can be any combinations of the remaining seven numbers. They can be $7!$ ways for each number in the above list.
So we have a total of $7! \times 16$ possibilities.
Probability is obtained by dividing favourable number by total number.
That is, the probability that a \[9\] digit number formed by $1,2,3,...,9$ without repetition is divisible by $36$ is obtained by dividing number of \[9\] digit numbers formed by $1,2,3,...,9$ without repetition which are divisible by $36$ by total number of \[9\] digit numbers formed by $1,2,3,...,9$ without repetition.
So we have, ${\text{Probability = }}\dfrac{{7! \times 16}}{{9!}}$
$ \Rightarrow {\text{Probability = }}\dfrac{{7! \times 16}}{{9 \times 8 \times 7!}}$
Cancelling $7!$ from numerator and denominator we have,
${\text{Probability = }}\dfrac{{16}}{{9 \times 8}} = \dfrac{2}{9}$
$\therefore $ The answer is option B.

Note: The important point here is that the digits are taken without repetition. If repetition was allowed, more words could have been made. Then the probability too will be changed.