What is the probability that the number so obtained is a multiple of $5$?
I could figure out that the total number of possible outcomes are $9 \times 9!$. How do I fix $0$ or $5$ at the end and also we'll have to exclude the number $10^9$ since it's not greater than itself!
On
The probability $p$ in question can be computed by conditioning on the first digit: $$\eqalign{p&=P\bigl[1^{\rm st}=5\bigr]\cdot P\bigl[{\rm last}=0\bigm|1^{\rm st}=5\bigr]+P\bigl[1^{\rm st}\ne5\bigr]\cdot P\bigl[{\rm last}\in\{0,5\}\bigm|1^{\rm st}\ne5\bigr]\cr &={1\over9}\cdot{1\over9}+{8\over9}\cdot{2\over9}={17\over81}\ .\cr}$$
You are correct that the number of possible outcomes is $9 \cdot 9!$ since the leading digit can be chosen in $9$ ways, and the remaining nine digits can be arranged in $9!$ ways. Since the leading digit cannot be zero, the number is greater than $10^9 = 1,000,000,000$.
If the number if a multiple of $5$, its units digit is $0$ or $5$. We consider these cases separately.
The units digit is $0$: The first nine digits of the number can be arranged in $9!$ ways.
The units digit is $5$: Since $0$ cannot be the leading digit, the leading digit can be selected in $8$ ways. The remaining eight digits can be arranged in $8!$ ways. Hence, there are $8 \cdot 8!$ such numbers.
The probability that a $10$-digit positive integer formed by permuting the digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ is a multiple of $5$ is $$\frac{9! + 8 \cdot 8!}{9 \cdot 9!}$$