A restaurant serves $5$ spicy dishes and $45$ regular dishes. A group of 30 people each orders dishes, with at most $2$ spicy dishes ordered. How many possible ways of ordering are there if
a) each dish has to be different?
b) a dish can be ordered multiple times (but still at most $2$ spicy dishes)?
I know my original answers were wrong. While going over the problem, I think I realized what my mistake was, but want to check. Originally, I thought:
a) Let $A_0$ be the case that $0$ spicy dishes are ordered, $A_1$ that $1$ spicy dish is ordered, and $A_2$ that $2$ spicy dishes are ordered.
Case $A_0$: Choose $1$ person from $30$ to order $1$ from $45$ available dishes. Choose $2$nd person from remaining $29$ to choose one of $44$ dishes. Continuing, we get $$\binom{30}{1}\binom{45}{1}\binom{29}{1}\binom{44}{1} \ldots = 30! \frac{45!}{(45-30)!}$$ Since we only care about final arrangement (not who orders first, second,..), answer simplifies to ${45}\choose{30}$
case $A_1$: Just have to choose $1$ person from $30$ to order one of $5$ spicy dishes. Remaining part is as in part a, except with $29$ people: $$30\cdot 5\cdot \binom{45}{29}$$
case $A_2$: choose $2$ people to order spicy dishes, choose $2$ of $5$ spicy dishes, and as before, we get: $$30\cdot 29 \cdot 5\cdot 4\cdot \binom{45}{28}$$
Was my mistake considering who will order the spicy dishes in $A_1$ and $A_2$; Should the answer have just been $$A_0 + A_1 + A_2 = \binom{45}{30} + \binom{5}{1}\binom{45}{29} +\binom{5}{2}\binom{45}{28}?$$
B) Let $B_0$ be $0$ spicy dishes ordered, each dish can be ordered as many times as possible (except at most $2$ spicy dishes). Similar for $B_1$ and $B_2$.
-case $B_0$: $45$ options for each of $30$ people, so $|B_0| = 45^{30}$
-case $B_1$: Someone chooses $1$ spicy dish, $45$ options for each of $29$ people, so $|B_1|=5 \cdot 45^{30}$ (I originally put $30 \cdot 5 \cdot 45^{29}$ but I know this was wrong)
-case $B_2$: as before, $|B_0| = 5^2 \cdot 45^{28}$ (again, originally had $30 \cdot 29 \cdot 5^2 \cdot 45^{28}$)
The number of ways a subset of exactly $k$ of the $5$ spicy dishes and exactly $30 - k$ of the $45$ normal dishes can be selected is $$\binom{5}{k}\binom{45}{30 - k}$$ Thus, the number of selections of $30$ different dishes that contain at most two spicy dishes is $$\sum_{k = 0}^{2} \binom{5}{k}\binom{45}{30 - k} = \binom{5}{0}\binom{45}{30} + \binom{5}{1}\binom{45}{29} + \binom{5}{2}\binom{45}{28}$$ However, it matters which customer receives which dish. Therefore, we must multiply by the $30!$ ways of assigning the selected dishes to customers. Hence, the number of possible orders that can be placed is $$30!\sum_{k = 0}^{2} \binom{5}{k}\binom{45}{30 - k} = 30!\left[\binom{5}{0}\binom{45}{30} + \binom{5}{1}\binom{45}{29} + \binom{5}{2}\binom{45}{28}\right]$$
There are $\binom{30}{k}$ ways for exactly $k$ of the customers to order a spicy dish. Each of those $k$ customers has $5$ choices. Each of the remaining $30 - k$ customers has $45$ choices. Hence, there are $$\binom{30}{k}5^k45^{30 - k}$$ orders in which exactly $k$ of the customers orders a spicy dish. Since at most two customers order a spicy dish, the number of possible orders is $$\sum_{k = 0}^{2} \binom{30}{k}5^k45^{30 - k} = \binom{30}{0}5^045^{30} + \binom{30}{1}5^145^{29} + \binom{30}{2}5^245^{28}$$