There are 40 possible combinations of numbers including one, two and three digit prime numbers.
The total quantity of numbers formed by a certain quantity of digits is equal to the total number of permutations, that is the number of combinations where order is taken into account. The permutation formula is presented below:
[tex]_n\mathbb{P}_k = \frac{n!}{(n-k)!}[/tex] (1)
Where:
- [tex]n[/tex] - Total quantity of available digits.
- [tex]k[/tex] - Used quantity of available digits.
Now we determine the total number of one, two and three digit prime numbers is:
[tex]x = _4\mathbb{P}_{1} + _4\mathbb{P}_{2} + _4\mathbb{P}_{3}[/tex]
[tex]x = \frac{4!}{(4-1)!}+\frac{4!}{(4-2)!} + \frac{4!}{(4-3)!}[/tex]
[tex]x = 4 + 12 + 24[/tex]
[tex]x = 40[/tex]
There are 40 possible combinations of numbers including one, two and three digit prime numbers.
We kindly invite to check this question on permutations: https://brainly.com/question/14767366
