How many 5 digit numbers are divisible by 4 can be formed?

Hint: We will use the concepts of permutation and combination to solve this question. The basic combination formula $\dfrac{n!}{r!\left( n-r \right)!}$ is used to solve. Also the divisibility rule for the number to be divisible by $25$ will be used to get the desired answer.

Complete step by step answer:
We have been given the digits $0,1,2,3,4,5$.
We have to find the number of 5 digit numbers divisible by 25 that can be formed using digits $0,1,2,3,4,5$.
Now, we know that for a number to be divisible by $25$ the last two digits of the number must be either $25$ or $50$.
So, if the last two digits are $2$ and $5$and the first digit cannot be zero so we have only three choices i.e ${}^{3}{{C}_{1}}$ and remaining two places can be filled by the remaining three numbers in ${}^{3}{{P}_{2}}$ ways.
So we have total ${}^{3}{{C}_{1}}\times {}^{3}{{P}_{2}}$ ways.
Now, solving these we get
\[\begin{align}
  & \Rightarrow {}^{4}{{C}_{3}}=\dfrac{4!}{3!\left( 4-3 \right)!} \\
 & \Rightarrow {}^{4}{{C}_{3}}=\dfrac{4\times 3!}{3!1!} \\
 & \Rightarrow {}^{4}{{C}_{3}}=4 \\
\end{align}\]
Now, the total possible numbers will be $18+4=22$
So, the number of 5 digit numbers divisible by 25 that can be formed using digits $0,1,2,3,4,5$ are $22$.

Option C is the correct answer.

Note:
The key concept to solve this question is by fixing the last two digits of the number which are divisible by 25 by using the divisibility rule and rearrange the other three digits by using permutation and combination rules.

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