Question

How many eight-digit numbers can be formed if the leading digit cannot be zero and the last number cannot be one.

Answer 1

Since the first one can't be 0, there are 9 possibilities for it.

Also, since the last one can't be 1, there is also 9 possibilities for it.

The remainder 6 digits have 10 possibilities.

Since the order does matter, because the order of the digits make the numbers different, the number of possibilities can be calculated by multiplying the number of possibilities of each digit.

So, we have:

`\begin{gathered} 9\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot9 \\ 9\cdot9\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10 \\ 81\cdot1000000 \\ 81000000 \end{gathered}`

So, there are 81000000 possibilities.