How many of them?

The five numbers can be arranged in $5!$ times $= 120.$

And If zero is in front it is not considered as a five digit number, So keeping zero at first, The remaining four numbers can be arranged by $4! = 24.$

Therefore, Number of five digit numbers made using the numbers, $= 120 - 24 = \color{#D61F06}{\boxed{96}}.$

The number of possibilities of ten-thousandth digit = $\color{#20A900}{4}$ (0 is excluded because then the number does not become a 5-digit number)
The number of possibilities of thousandth digit = $\color{#3D99F6}{4}$ (one digit gone but 0 included)
The number of possibilities of hundredth digit = $\color{magenta}{3}$ (two digits gone)
The number of possibilities of tenth digit = $\color{#CEBB00}{2}$ (three digits gone)
The number of possibilities of units digit = $\color{#E81990}{1}$ (four digits gone)

$\therefore$ Number of 5-digit numbers required = $\color{#20A900}{4}×\color{#3D99F6}{4}×\color{magenta}{3}×\color{#CEBB00}{2}×\color{#E81990}{1} = \color{#D61F06}{\boxed{96}}$

Answer is (4!)×4.