# Ways of $30$ people ordering from $5$ spicy dishes and $45$ normal dishes

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 $$30 \cdot 29 \cdot 5^2 \cdot 45^{28}$$)

• part A is correct i.e $\binom{45}{30} \times \binom{5}{0}+\binom{45}{29} \times \binom{5}{1}+\binom{45}{28} \times \binom{5}{2}$ .I think Part B should be $45^{30}+46^{30}+47^{30}$ Jul 5, 2018 at 9:48
• But for part B, if 2 people order 2 spicy dishes (which could be the same spicy dish), then none of the other 28 people can order a spicy dish. The condition of ordering at most 2 spicy dishes between the 30 people still applies. Jul 5, 2018 at 9:56
• @laura Part (a) is not correct since it matters which customer orders which dish. Your answer for part (b) is incorrect since the customers who order a spicy dish have only five choices and the customers who do not order a spicy dish have only $45$ choices. Jul 5, 2018 at 10:04
• To produce $\binom{n}{k}$, type \binom{n}{k} when you are in math mode. Also, you can enclose the entire mathematical expression within dollar signs for an inline equation and double dollar signs for a displayed equation. Jul 5, 2018 at 10:16

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 each dish is different?
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 $$30!\sum_{k = 0}^{2} \binom{5}{k}\binom{45}{30 - k} = 30!\left[\binom{5}{0}\binom{45}{30} + \binom{5}{1}\binom{45}{29} + \binom{5}{2}\binom{45}{28}\right]$$
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 dish can be ordered multiple times?
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 $$\sum_{k = 0}^{2} \binom{30}{k}5^k45^{30 - k} = \binom{30}{0}5^045^{30} + \binom{30}{1}5^145^{29} + \binom{30}{2}5^245^{28}$$