Question

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

Answer 1

Hint: We use the fact that a number of divisible 36 has to be divisible by both 4 and 9. We see that since sum of the digits from 1 to 9 is 36, the nine-digit number will always be divisible by 9. So we only find the 9 digits number divisible by 4 as a number of favorable outcomes $ n\left( A \right) $. We find the number of total outcomes as the number of ways we can arrange 9 digits without repeating as $ n\left( S \right) $. The required probability is $ P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)} $ .\[\]

Complete step by step answer:
We know from the definition of probability that if there is $ n\left( A \right) $ number of ways of event $ A $ occurring (or number of favorable outcomes) and $ n\left( S \right) $ is the size of the sample space (number of all possible outcomes) then the probability of the event $ A $ occurring is given by
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}\]
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
\[n\left( S \right)=9!\]
Let us denote the event of getting a number divisible by 36 as $ A $. 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 last two digits is divisible. So the possible numbers than can be formed last two digits are
\[\begin{align}
  & 12,32,52,72,92 \\
 & 24,54,84, \\
 & 16,36,56,76,96, \\
 & 28,48,68 \\
\end{align}\]
So we have 16 combinations to fill the one and ten’s place of the 9digit number. We can arrange the rest 7 digits in $ 7! $ ways. So the number 9-digit numbers that is divisible by 4 is $ 16\times 7! $ . So the number 9-digit numbers that is divisible by 4 is the number of favourable out comes as
\[n\left( A \right)=16\times 7!\]
So the required probability is
\[P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{16\times 7!}{9!}=\dfrac{16}{8\times 9}=\dfrac{2}{9}\]
We are given in the question that $ p=P\left( A \right)=\dfrac{2}{9} $ . So the requires result is $ 9p=9\times \dfrac{2}{9}=2 $ .\[\]

Note:
 We note that if a number is divisible by $ n=a\times b $ then it has to be divisible by $ a $ and $ b $ only when $ a,b $ are relative prime. We note that the equation presumes that b random order means there will not be any repetition of digits. If the digits will be repeated then the number of 9-digit numbers is $ {{9}^{9}} $.