Counting flower and committee questions $1$) You want dozen roses. The florist has white, pink, red, and violet roses. How many possible ways could you make the order?
$2$) There are $35$ men and $15$ women. Committee needs to have four men and four women, the chair will be a woman and secretary a man. How many ways can the committee be filled?
My attempt:
$1$) For the first one, I am not sure which one is right answer:  $\dfrac{4^{12} -4}{12} + 4$ or $\dbinom{12+4-1}{12}$.
$2$) For the second one, my answer is $$\dbinom{35}4 \times \dbinom{15}4 \times \dbinom{35-4}1 \times \dbinom{15-4}1$$
Am I thinking this correctly?
 A: For the first problem, note that the order of the flowers are not important, i.e., there is only one possible choice of $4$ red, $2$ white, $1$ pink and $5$ violet flowers. Hence, you need to count the number of non-negative solutions solutions to
$$w+p+r+v = 12$$
The number of non-negative integer solutions to $\displaystyle \sum_{i=1}^{n} a_i = N$ is given by $C(N+n-1,n-1)$. Look up stars and bars.

For the second problem, first choose the chair. There are $15$ choices. Then the secretary. This has $35$ choices. For the remaining members, you need to choose $3$ men out of $34$ and $3$ women out of $14$ women. Hence, the number of ways is
$$15 \times 35 \times \dbinom{35-1}3 \times \dbinom{15-1}3$$
For the second one, your line of thought is correct. However, I assume that the committee of $4$ men and $4$ women include the chairman and secretary.
Also, note that there are multiple ways to count the second one. You can first select $4$ men and $4$ women and then choose the chairman and secretary, i.e., it will be $$\dbinom{35}4 \times \dbinom{15}4 \times \dbinom{4}1 \times \dbinom{4}1$$
Alternately, you can choose the $3$ men and $3$ women who will be the committee member (not the secretary or chairman) and then choose the secretary and chairman, which will give us $$\dbinom{35}3 \times \dbinom{15}3 \times \dbinom{35-3}1 \times \dbinom{15-3}1$$
All these will give you the same answer.
