In how many ways can the digits of $45025$ be arranged to form a $5$-digit number such that it is divisible by $5$? 
In how many ways can the digits of $45025$ be arranged to form a $5$-digit number such that it is divisible by $5$?

I got $$\frac{4 \cdot 4 \cdot 3 \cdot 2}{2!} = 48$$ different arrangements for how many $5$-digit numbers you can make with the digits of $45025$. The problem is I'm not quite sure how to tackle the second part. I know for a number to be divisible by 5 it has to end with either a $0$ or $5$, but I'm not sure how to find how many combinations $45025$ can make so that it has $0$ or $5$ at the end.
 A: You have to make cases.
CASE 1:
No.of ways when unit digit is $5$ =$4!=24$
No.of ways when unit digit is $5$ and leading digit is $0$ =$3!=6$
Thus,valid cases when unit digit is $5=24-6=18$
CASE 2:
No.of ways when unit digit is $0$ =$4!/2!=12$
Thus,total no. of ways is $18+12=30$
.
A: Method 1:  We consider cases, depending on whether the units digit is $0$ or $5$.
Case 1:  The units digit is $0$.
We have to arrange the digits $2, 4, 5, 5$ in the first four positions.  Choose two of those four positions for the $5$s, then arrange the remaining two distinct digits in the remaining two positions, which can be done in
$$\binom{4}{2}2!$$
ways.
Case 2:  The units digit is $5$.
We have to arrange the digits $0, 2, 4, 5$ in the first four positions.  There are three possible positions for the $0$ since it cannot be the leading digit.  Once it is placed, arrange the remaining three distinct digits in the remaining three positions. There are
$$3 \cdot 3!$$
such numbers.
Total:  Since the two cases are mutually exclusive and exhaustive, the number of ways the digits of the number $45025$ can be permuted to form a five-digit positive integer that is divisible by $5$ is
$$\binom{4}{2}2! +  3 \cdot 3! = 30$$
Method 2: We subtract those integers formed by permuting the digits of $45025$ which are not divisible by $5$ from the $48$ total permutations you found.
Those integers have a $2$ or $4$ in the units digit.  Choose whether $2$ or $4$ is the units digit.  Since $0$ cannot be the leading digit, choose which of the three middle positions is occupied by the $0$.  Choose which of the remaining three positions is occupied by whichever of the digits $2$ or $4$ we have not already used.  The two $5$s must occupy the remaining two positions.  Hence, there are
$$2 \cdot 3 \cdot 3 = 18$$
five-digit positive integers that can be formed by permuting the digits of $45025$ which are not divisible by $5$, leaving
$$48 - 18 = 30$$
five-digit positive integers that can can be formed by permuting the digits of $45025$ which are divisible by $5$.
