Question

How many 6-digit numbers can be formed using the digits 1,2,3,4,5,6 which are divisible by 4 and digits are not repeated?
A. 122
B. 192
C. 164
D. 155

Answer 1

Hint: We have to find the total number of 6 digit numbers using the digits 1,2,3,4,5,6 which are divisible by 4 and none of the digits are repeated. For this, we will use the divisibility test of ‘4’ in which the last two digits of any number should be a multiple of 4. Then we will count in how many total ways the digits 1,2,3,4,5,6 can be arranged so that the last two digits come out to be a multiple of 4. These total number of ways will be our answer.

Complete step-by-step solution:
Now, from 1,2,3,4,5,6, multiples of 4 are:
12,16,24,32,36,52,56,64
Also, we have been given 1,2,3,4,5,6 to form the required number therefore there are a total of 6 digits available to us.
Therefore, there are a total of 8 multiples of 4 which can be formed from the given digits without repetition.
For the 6 digit number to be divisible by 4, the last 2 digits of the number should be a multiple of 4, i.e., it should be any of the 8 multiples stated above.
Therefore, there are a total of 8 ways in which the last two digits of the 6 digit number can be filled.
For the first digit of the 6 digit number,
Multiples of 4 have been formed by two distinct digits out of the given 6 digits therefore there are a total of $'6-2=4'$ ways in which the first digit can be filled.
Now for the second digit, there will be $'6-3=3'$ ways available to fill it as 2 numbers are used by the last 2 digits of the number and one is used by the first digit.
Similarly, there will be $'6-4=2'$ ways available for the third digit and $'6-5=1'$ ways available for the fourth digit.
Thus, the total number of 6 digit numbers that can be formed by the digits 1,2,3,4,5,6 without repetition such that the number is divisible by 4 are given as:
$\begin{align}
  & \Rightarrow 4\times 3\times 2\times 1\times 8 \\
 & \Rightarrow 192 \\
\end{align}$
Therefore, there are a total of 192 numbers which can be formed using the digits 1,2,3,4,5,6 without repetition such that the number is divisible by 4.
Hence, option (B) is the correct option.

Note: This question can also be done by the following method:
After establishing that the last two digits can be filled in a total of 8 ways, there will be $'6-2=4'$ digits left which will be available to us to fill the first four digits of the number.
As long as the last two digits of the number are a multiple of 4, the first 4 digits can be filled in any way.
Therefore, the total number of ways in which the first 4 digits will be filled in will be equal to $^{4}{{P}_{4}}$ as out of the 4 available digits, 4 digits have to be selected and arranged.
Therefore, the number of ways in which the required number of 6 digit number can be formed is:
$\begin{align}
  & {{\Rightarrow }^{4}}{{P}_{4}}\times 8 \\
 & \Rightarrow \dfrac{4!}{\left( 4-4 \right)!}\times 8 \\
 & \Rightarrow \dfrac{4!}{0!}\times 8 \\
 & \Rightarrow 4!\times 8 \\
 & \Rightarrow 24\times 8 \\
 & \Rightarrow 192 \\
\end{align}$