A bakery makes 4 varieties of cookies (chocolate chip, peanut butter, ginger snaps and oatmeal-raisin).
(a) How different trays of a dozen cookies can the bakery offer?
(b) How many of those trays will contain at least one peanut butter cookie ?
For part $a$ we have $4$ distinct types of cookies and we want to select 12 so the answer would be $${4 + 12 - 1 \choose 12 } = {15 \choose 12}$$
Now for part $b$ I am thinking we could think about 12 jars and we put one peanut cutter in one of those jars and the remaining 11 jars we are free to put whatever we want so we would have $${11 + 4 -1 \choose 11 } = {14 \choose 11}$$
Is that correct reasoning ?
The phrasing of the question you are most comfortable with varies from person to person. The form of this question that I learned first and feel most comfortable with is in the following way:
How many non-negative integer solutions exist for the following system:
$\begin{cases} x_1+x_2+x_3+x_4=12\\ 0\leq x_1\\ 0\leq x_2\\ 0\leq x_3\\ 0\leq x_4\end{cases}$
where here, $x_1$ represents the number of chocolate chip cookies, $x_2$ represents the number of peanut butter cookies, etc...
The second part putting the condition that "there is at least one peanut butter cookie" is like changing the system to the following:
$\begin{cases} x_1+x_2+x_3+x_4=12\\ 0\leq x_1\\ 1\leq x_2\\ 0\leq x_3\\ 0\leq x_4\end{cases}$
We can make a change of variable here, setting $y_2=x_2-1$ and $y_i=x_i$ for the rest, we have then that $y_1+y_2+y_3+y_4=x_1+(x_2-1)+x_3+x_4=(x_1+x_2+x_3+x_4)-1=12-1=11$, yielding the system
$\begin{cases} y_1+y_2+y_3+y_4=11\\ 0\leq y_1\\ 0\leq y_2\\ 0\leq y_3\\ 0\leq y_4\end{cases}$