Calculate how many integers between $0$ to $9999$ that has the digits $2,5,8$. That is integers that has each of the three numbers at least once.
This is similar to How many numbers between $0$ and $9999$ have either of the digits $2,5,8$ at least once - Check my answer but mine is quite different.
My question is how do you calculate it by the 'straight forward' method ?
I have a proposed solution:
If the last digit is not one of the three then we have: $3!\cdot8$ possibilities.
If the last digit is one of them, we have $(3\cdot3!)/2$ we divide by two for symmetry (2258 is counted twice for example).
In total we get: 57 such integers.