So, I do understand the concept of permutations, but I can't figure out how the formula for finding permutations even applies to this. The author solves it as follows, but he does not make any explicit comments as to why he did what he did:
\begin{align*}&= 3 (^3P_3)+(^4P_4)\\ &= 3 (3! + 4!) \\ &=3 ( 6+24)\\ &=3(30)\\ &= 90 \end{align*}
Would appreciate any explanation of what is going on here.
Presumably we are looking for the number of numbers greater than $23000$ yada yada.
So, we can break into two cases.
In the first case, the leading digit is a $2$. If that were the case, then the following digit must not be a $1$, otherwise it would be a number smaller than $23000$. Then, from there, fill in the remaining digits in such a way as so that no digit is reused. There are $3$ choices available for the second digit, $3$ remaining choices available for the third, $2$ remaining choices available for the fourth, and $1$ remaining choice available for the fifth.
This gives us a total of $3\times 3\times 2\times 1$ possible numbers where the first digit is a $2$ such that the number is greater than $23000$, the available digits are $1,2,3,5,6$ and no digit is repeated.
The remaining case is similar. The first digit may not be a $1$, and further the first digit may not be a $2$ (otherwise we would be back in the first case or the number would be too small). That leaves $3$ choices available remaining for the first digit. Then, fill in the remaining digits. There are $4$ choices remaining available for the second digit and so on...
This gives $3\times 4\times 3\times 2\times 1$ possible numbers in the second case.
Adding these together gives the final result.
