# How many possible orders are there?

A tapas bar serves 15 dishes, of which 7 are vegetarian, 4 are fish and 4 are meat. A table of customers decides to order 8 dishes, possibly including repetitions.

a) Calculate the number of possible dish combinations.

b) The customers decide to order 3 vegetarian dishes, 3 fish and 2 meat. Calculate the number of possible orders.

Progress. For a) I think that the answer would be $15^8$ as this would be the number of different ordered sequences of 8 elements from the 15 possible dishes.

• Welcome to Mathematics! For questions of this sort, it is customary to also mention what you know about the problem, what work you have put into the problem, as well as what your thoughts are toward the overall solution. That way, we can deliver the correct level of help.
– Ken
May 24, 2015 at 13:45
• Welcome to math stack exchange! May 24, 2015 at 13:45
• Are all vegetarian dishes considered the same? Or is it different to order 8 times the first vegetarian dish vs. 8 times the second vegetarian dish? May 24, 2015 at 13:47
• For a) I think that the answer would be 15^8 as this would be the number of different ordered sequences of 8 elements from the 15 possible dishes. May 24, 2015 at 13:50
• Indeed $15^8$ is the number of sequences that are possible. I would have assumed, however, that we do not want to count "seven of the first fish dish and one of the first meat dish" as a different selection of dishes than "one of the first meat dish and seven of the first fish dish." May 24, 2015 at 14:21

a) First, we note that repetition is allowed, and the order in which we order the dishes is unimportant. Therefore, we use the formula ${k+n-1 \choose k}$. In this case, $n=15$ and $k=8$. So, the number of possible combinations of dishes is $${8+15-1 \choose 8} = {22 \choose 8} = 319770.$$
b) We will need to apply the above formula three times. For the vegetarian dishes, we have $n_V = 7$ and $k_V=3$. For the fish dishes, we have $n_F = 4$ and $k_F=3$. For the meat dishes, we have $n_M = 4$ and $k_M=2$. So, the number of possible combinations of dishes under these conditions is $${k_V+n_V-1 \choose k_V}{k_F+n_F-1 \choose k_F}{k_M+n_M-1 \choose k_M} = {9 \choose 3}{6 \choose 3}{5 \choose 2}=16800.$$
a) The given that all dishes are different, then $$N_{orders}=15^8$$
b) In this case, we can calculate how many different orders for each type and then combine them: $$N_{orders} = 7^3 \times 4^3 \times 4^2$$