Originally Posted by

**Soroban** Hello, william!

I will assume that the number may __not__ begin with 0 (zero).

The first digit cannot be zero: 9 choices.

The second digit can be any of the other 9 other digits.

The third digit can be any of the other 8 digits.

The fourth digit can be any of the other 7 digits.

The fifth digit can be any of the other 6 digits.

The sixth digit can be any of the other 5 digits.

Answer: .$\displaystyle 9 \times 9 \times 8 \times 7 \times 6 \times 5 \:=\:130,\!080$

The first digit has 9 choices.

Each of the other five digits has 10 choices.

Answer: .$\displaystyle 9 \times 10^5 \:=\:900,\!000$

In part (b), we saw that there are $\displaystyle 900,\!000$ possible numbers.

Consider the numbers that have **no** 5's.

The first digit has 8 choices (not 0 and not 5).

Each of the other five digits has 9 choices.

Hence, there are: .$\displaystyle 8 \times 9^5 \:=\:472,392$ numbers with no 5's.

Answer: .$\displaystyle 900,\!000 - 472,\!392 \:=\:427,608$

The number is $\displaystyle \geq 600,\!000$

The first digit must be: 6, 7, 8, or 9 . . . 4 choices.

The number is even.

The last digit must be: 0, 2, 4, 6, 8 . . . 5 choices.

Each of the other four digits has 10 choices.

Answer: .$\displaystyle 4 \times 5 \times 10^4 \:=\:200,\!000$