Question

A ten digit number is formed using the digit 0 to 9, every digit being used exactly once. Find the probability that the number is divisible by 4.
$A.{\text{ }}\dfrac{{20}}{{81}} \\\ B.{\text{ }}\dfrac{{16}}{{81}} \\\ C.{\text{ }}\dfrac{{22}}{{81}} \\\ D.{\text{ }}\dfrac{{25}}{{81}} \\\$

Answer 1

Hint- Use the concept of factorial and permutation in order to count the number of required numbers favorable and total number of numbers and finally use the basic definition of probability to get the final answer. Use the fact that for a number to be divisible by 4 its last two digits must be divisible by 4.

Complete step-by-step answer:
Given that:
We need to construct a 10 digit number using digits $0,1,2,3.....9$ exactly once.
As we know that if we need to construct number using n digits exactly once then total number of such numbers is $= n! - \left( {n - 1} \right)!$
Total number of such 10 digit numbers are $= 10! - 9!$
Since zero cannot be the first term.
Now to find the numbers divisible by 4.
We know for any number to be divisible by 4 the last two digits of that number must be divisible by 4.
Hence the last two digits must be a multiple of 4.
From 01 to 99 we have 24 multiples of 4. They are:
\left\\{ {04,08,12,16,20,...........,88,92,96} \right\\}
So whenever these numbers appear as last term, the number will be divisible by 4 altogether.
$\Rightarrow$ Out of these 24 number 2 numbers are \left\\{ {44,88} \right\\} which has repeating digits. So these won’t be considered.
Now out of rest 22 numbers 6 numbers have 0 digit with them. They are:
\left\\{ {04,08,20,40,60,80} \right\\}
$\Rightarrow$ So number of number with these 6 numbers as last 2 digits are: $8!$ for each
i.e. in all $6 \times \left( 8 \right)!$ number can be formed.
$\Rightarrow$ Now, from the rest of 16 numbers as the last 2 digits.
Total numbers to be formed by each one $= 8! - 7!$
So, altogether $16\left( {8! - 7!} \right)$ can be formed by them.
$\therefore$ Total number of numbers divisible by 4 $= 6\left( 8 \right)! + 16\left( {8! - 7!} \right)$
Therefore the required probability is
$= \dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}} \\\ = \dfrac{{6\left( 8 \right)! + 16\left( {8! - 7!} \right)}}{{10! - 9!}} \\\$
To solve the factorial we will take $7!$ common from numerator and denominator.

$$= \dfrac{{6\left( 8 \right)! + 16\left( {8! - 7!} \right)}}{{10! - 9!}} \\\ = \dfrac{{6 \times 8 \times \left( 7 \right)! + 16\left( {8 \times 7! - 7!} \right)}}{{\left( {10 \times 9 \times 8} \right) \times 7! - \left( {9 \times 8} \right) \times 7!}} \\\ = \dfrac{{\left( 7 \right)!\left( {6 \times 8 + 16\left( {8 - 1} \right)} \right)}}{{\left( 7 \right)!\left( {\left( {10 \times 9 \times 8} \right) - \left( {9 \times 8} \right)} \right)}} \\\ = \dfrac{{\left( {6 \times 8 + 16\left( {8 - 1} \right)} \right)}}{{\left( {\left( {10 \times 9 \times 8} \right) - \left( {9 \times 8} \right)} \right)}} \\\ = \dfrac{{48 + 112}}{{720 - 72}} \\\ = \dfrac{{160}}{{648}} \\\ = \dfrac{{20}}{{81}} \\\$$

Hence, the probability of a 10 digit unique number to be divisible by 4 is $\dfrac{{20}}{{81}}$ .
So option A is the right option.

Note- This question required a combined effort of concepts of probability, factorial and permutation. The concept of digits in order to find if the number is divisible or not by a digit is very easy but still must be remembered. The concept of factorial must be remembered and if there is a huge calculation involving factorial of larger number then never try to find the factorial by direct multiplication, rather try to cancel out some numbers as done above.