Q:

Number of ways to form a 6-digit sequence with these constraints.

Doing some combinatorics exercises, I am having doubts with the last question in this - can you tell if it makes sense?

In how many ways can you form a 6-digit sequence if it must begin with
  a digit different than $0$ and end with an even digit. Repetitions are
  ok.

$$9\cdot \frac{10!}{(10 - 4)!}\cdot5$$

What if such sequence contains exactly two zeroes and four nines?

Since it must begin with a digit different than $0$ and end with an even number, we already know that the sequence is of the form
$$9 \ \_ \ \_ \ \_ \ \_ \ 0 $$
Hence we need only to decide the $4$-digit sequence, composed of one zero and three nines.
$$4!$$

What if such sequence contains, at most, $5$ zeroes?

$$\sum_{i = 1}^{5}\left( 9\cdot \frac{9!}{(9 - 6 - i)!}\cdot 5^{i-1}\right)$$

A:

In the first question: $9\cdot 10^4\cdot 5$ since you have $9$ choices for the first digit, $10$ for digits $2$ through $5$ and $5$ choices for the last digit.
In the second question I agree with the fact that the sequence is of the form $\text{9 _ _ _ _ 0}$ However there are only 4 ways to select which of the spots in the middle is the zero.
In the last question: Any 6 digit sequence not starting with zero has at most five zeroes, so the answer is the same as the first question.

