A croissant shop has plain, cherry, chocolate, almond, apple, and broccoli croissants. a) How many ways are there to choose a dozen croissants?

b) How many ways are there to choose three dozen croissants?

c) How many ways are there to choose two dozen croissants with at least two of each kind?

d) How many ways are there to choose two dozen croissants with no more than two broccoli croissants?

How I thought to do the problem:

a) $6^{12}$ because you can have repetition.

b) $6^{36}$ - same argument as a.

c) If you have to pick at least two of each kind, then you subtract a dozen croissants that you are allowed to choose because you have six different varieties. This gives $6^{12}$.

d) If you have two dozen croissants with no more than two broccoli croissants, then you can first take all the possibilities with zero broccoli croissants ($5^{26}$) + having one broccoli croissant ($5^{25}$) + having two broccoli croissants ($5^{24}$).

Am I correct in my thinking?
Given Question is of the form "selecting indistinguishable objects into distinguishable boxes" and the formula is given by  $$\binom{n+r-1}{r}$$ where $n=$number of distinguishable boxes and $r=$number of indistinguishable object.

a) $$\binom{6+12-1}{12} =6188$$
b  $$\binom{6+36-1}{36} =749398$$
c) $$\binom{6+(24-12)-1}{12} =6188$$
d) $$\underbrace{\binom{5+(24-0)-1}{24}}_\text{zero broccoli croissants} + \underbrace{\binom{5+(24-1)-1}{23}}_\text{one broccoli croissants} +  \underbrace{\binom{5+(24-2)-1}{22}}_\text{two broccoli croissants} =20475+17550+14950=52975$$
Given Question is of the form "selecting indistinguishable objects into distinguishable boxes" and the formula is given by  $$\binom{n+r-1}{r}$$ where $n=$number of distinguishable boxes and $r=$number of indistinguishable object.

a) $$\binom{6+12-1}{12} =6188$$
b  $$\binom{6+36-1}{36} =749398$$
c) $$\binom{6+(24-12)-1}{12} =6188$$
d) $$\underbrace{\binom{5+(24-0)-1}{24}}_\text{zero broccoli croissants} + \underbrace{\binom{5+(24-1)-1}{23}}_\text{one broccoli croissants} +  \underbrace{\binom{5+(24-2)-1}{22}}_\text{two broccoli croissants} =20475+17550+14950=52975$$