I have an assignment and just wanted to verify if my answer is correct.
Possibilities:
main courses: cheeseburger, fish
side dishes: lentils, fries
desserts: chocolate cake, vanilla ice, fruit salad
How many ways are there for a guest to create a menu. You can choose 1 dish of the main courses, 3 of the side dishes, 2 of the desserts.
My solution would be this:
$$
{2 \choose 1} \cdot \frac{5!}{2!} \cdot {3 \choose 2} = 360
$$
-Two choices for the main course : burger or fish.
-$4$ choices for the side dish : Either you pick no fries, one fries, two fries or three fries, the rest is completed by lentils.
-${3 \choose 2}$ of desserts.
The result is $$2*4*{3 \choose 2}$$