Good evening everyone. I'm trying to solve this math problem. How many 5 digit numbers can we create using only these numbers $(0; 3; 6; 7; 8)$. Created numbers cannot be divided by $5$ and the given numbers (the ones that we use to create the 5 digit number) can't repeat.
I know that correct answer is $72$, but I'm not sure how to get it. Maybe someone could help me with that? Thank you in advance
On
The number of ways to arrange $5$ digits total is $5! = 120$
We need to consider 2 other cases to meet the criteria for the question
Considering $0$ as the first digit: the number of ways to arrange $4$ digits total is $4!=24$
Considering $0$ as the last digit: the number of ways to arrange $4$ digits total is $4!=24$
We take the total number of combinations and subtract that combinations that are not allowed, so the total then becomes: $5! - 4! - 4! = 120 - 24 - 24 = 72$
There are $4$ ways to select the first digit (it cant be $0$).
After this there are $3$ ways to select the last digit (it cant be $0$ either).
After this there are $3!$ ways to select the numbers in the middle.
So the answer is $4\times 3\times 6=72$
Alternative solution:
Place the digit $0$ in any of $3$ positions and the rest in $4!$ ways.