Question is in title
My approach
We can approach this problem by counting the number of possible choices for each digit of the integer, and then multiplying the results together using the multiplication principle.
For the first digit, we have four choices $(0, 1, 2, or 4)$.
For the second digit, we have three choices (since we have used one of the four digits for the first digit).
For the third digit, we have two choices (since we have used two of the four digits for the first two digits).
For the fourth digit, we have one choice (since we have used three of the four digits for the first three digits).
Therefore, the total number of distinct positive integers that can be formed using $0$, $1$, $2$, and $4$, where each of these integers is used at most once, is:
$4 × 3 × 2 × 1 = 24$
So there are $24$ such integers. They are:
$0, 1, 2, 4, 10, 12, 14, 20, 21, 24, 41, 42, 40, 24, 41, 42, 40, 124, 120, 140, 214, 210, 240, 410, 412$
Is this right?
Best Answer
The number of 4-digit positive integers made using those digits: Pick the thousands-place digit (You have three nonzero options). Pick the hundreds-place digit (you have three remaining options). Pick the tens-place digit, and finally the ones-place digit.
The number of 3-digit positive integers using those digits: Pick the hundreds place digit (You have three nonzero options). Pick the tens-place digit (you have three remaining options). Pick the ones-place digit...
The number of 2-digit positive integers...
The number of 1-digit positive integers...
Add these totals together for the final answer.