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.
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:
$$
N_{orders} = 7^3 \times 4^3 \times 4^2
$$