How many numbers with $4$ distinct digits greater than $6000$ can be formed from $0,2,3,5,7,8,9$ ?
i try as follows : $$ \begin{array}{l}{\text { unit digit has } 6 \text { possibilities (except } 0 )} \\ {\text { tenth digit has } 6 \text { possibilities }} \\ {\text { hundreds digit has } 5 \text { possibilities }} \\ {\text { thousands digit has } 3 \text { possibilities }} \\ {\text { so answer is : } 6 \times 6 \times 5 \times 3=540} \\ {\text { is this true? any notes are welcomed }} \end{array} $$
You should start by choosing the thousands digit. Clearly there are three choices: $7,8$ and $9$.
Then leftover, you have $6$ choices for the hundreds unit, $5$ for the tens and $4$ for the units.
Therefore the answer should be $3\times 6\times 5\times 4 =360$.
Often, when dealing with probability questions, you'll want to satisfy the conditions first, then it'll make the rest of the problem easier. For this problem, a number to be greater than $6000$ requires the leading digit to be greater than $6$. Hence why there are only $3$ choices for the thousands digit. Once this is satisfied, the rest of the question is easy.