If it should not be greater than $4000$, should we consider $4000$ as well?
I used the following method:
Making four spots for our four digit number _ _ _ _, I can see that there would be only $3$ possibilities ($1,2,3$) at the thousands place, excluding $4$ as then the number would be greater than $4000$. The hundreds place can be filled in $5$ possible ways, and same for the tens and one place. Now using fundamental theorem of multiplication, I get $3\cdot5^3$ as my answer. Now the minimum possible value from my calculation would be $1000$ and hence I subtract from it. The maximum value I receive will be $3444$. Why should I add $1$ to it leaving all the cases from $3444$ to $3999$ just to include $4000$?
I am new here, please explain.
We wish to count all positive integers satisfying the inequalities $1000 < n \leq 4000$ composed only of the digits $0, 1, 2, 3, 4$.
Your count $3 \cdot 5^3$ counts all the integers composed of the digits $0, 1, 2, 3, 4$ that satisfy the inequalities $1000 \leq n < 4000$. However, we do not want to count $1000$, so we must subtract $1$, giving $3 \cdot 5^3 - 1$ integers satisfying the inequalities $1000 < n < 4000$. However, we have not counted $4000$, which satisfies the requirements that $1000 < n \leq 4000$ and uses only the digits $0, 4 \in \{0, 1, 2, 3, 4\}$. Therefore, we must add $1$ to the total, giving $3 \cdot 5^3 - 1 + 1 = 3 \cdot 5^3 = 3 \cdot 125 = 375$ positive integers that satisfy the stated conditions.
Note that every number satisfying $3445 \leq n \leq 3999$ contains a digit greater than $4$ in its decimal expansion, so they should not be counted.