For the below answer - Can we calculate the alternative solution (bolded one) using any formula for both the cases (with or without repetition)..?   Like first choose 1 cake for Steve out of 4 available options & then choose 1 cake for Ami out of 5 available options. I guess in this case order will be important and we could variation formula but it was not giving the right count.

Since Steve dislikes coconuts, our options are limited to 4 cakes. Then, we need to choose
two of the 4 remaining cakes, so 𝐶 =
𝑛!
𝑝!(𝑛−𝑝)!

4!
2!2!
= 6. If we can get two identical cakes,
then we have 𝐶̅=
(𝑛+𝑝−1)!
𝑝!(𝑛−1)!

5!
2!3!
= 10 options.
(Alternatively, we can get one Coconut cake and 1 other cake. That way Steve will still have
something else to eat. In that scenario, if we can have two identical cakes, then the only
option which we want to avoid is the double Coconut one. Thus, we take the 15 we got in
part b of a), and subtract 1, so we get 14 options.
Now, if we want to have 2 different cakes, we need to remove the double Cheesecake,
double Sacher, double Chiffon and double Carrot cake options. Therefore, there would be 10
different orders we could make.