Question

A four digit number is formed using the digits 0, 1, 2, 3, 4 without repetition. Find the probability that it is divisible by 4.

Answer 1

Hint : In the four-digit numbers the highest digit must not be zero. By keeping this condition in mind, find the total no. of four-digit numbers that can be formed using the digits 0, 1, 2, 3 and 4. A number is divisible by 4 only when the last two digits of the number is divisible by 4. So find the numbers which have the last two digits exactly divisible by 4.

Complete step-by-step answer :
We are given that a four digit number is formed using the digits 0, 1, 2, 3, 4 without repetition.
Using combinations here, because we have to select the digit from the 5 given digits
Four- digit number must have 4 digits, which means the leftmost digit should not be zero so it can be written in 4 ways, the next digit can be written in 4 ways, the next in 3 ways and the next in 2 ways.
So the number of four digit numbers can be formed using 0, 1, 2, 3, 4 is
$\Rightarrow {}^4{C_1} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1} \\\ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} \\\ = 4 \times 4 \times 3 \times 2 \\\ = 16 \times 6 \\\ = 96 \\\$
For a number to be divisible by 4, its last two digits must be divisible by 4. The last two digits must be a multiple of 4.
So using the digits 0, 1, 2, 3, 4, the possible multiples of 4 are 04, 40, 12, 20, 24, 32.
When any of the last two digits does not contain zero in it, then the leftmost digit can be written in only 2 ways as it cannot be zero.
So when the last two digits are 0, 4, then the first two places can be written in
$\Rightarrow {}^3{C_1} \times {}^2{C_1} = 3 \times 2 = 6ways$
When the last two digits are 4, 0, then the first two places can be written in
$\Rightarrow {}^3{C_1} \times {}^2{C_1} = 3 \times 2 = 6ways$
When the last two digits are 1, 2, then the first two places can be written in
$\Rightarrow {}^2{C_1} \times {}^2{C_1} = 2 \times 2 = 4ways$
When the last two digits are 2, 0, then the first two places can be written in
$\Rightarrow {}^3{C_1} \times {}^2{C_1} = 3 \times 2 = 6ways$
When the last two digits are 2, 4, then the first two places can be written in
$\Rightarrow {}^2{C_1} \times {}^2{C_1} = 2 \times 2 = 4ways$
When the last two digits are 3, 2, then the first two places can be written in
$\Rightarrow {}^2{C_1} \times {}^2{C_1} = 2 \times 2 = 4ways$
Total number of four-digit numbers that are divisible by 4 is $6 + 6 + 4 + 6 + 4 + 4 = 30$
The probability that the number is divisible by 4 is $\dfrac{{30}}{{96}} = \dfrac{5}{{16}}$

So, the correct answer is “Option C”.

Note : Combination is the selection of items from a collection. Use combinations when the order of the items that are selected does not matter and when the order matters use permutations.