Manhattan GRE confirms that, for last week's Math Beast Challenge problem, the correct answer is indeed (D)-- 210. MGRE's explanation:
Let’s look at the two possible combinations separately. The first type of combo is 1 entree, 2 side dishes, and 1 drink. When a customer orders this type, she makes several decisions:
Entree selection: choose 1 of 3, so 3 possibilities.
Side dish selection: choose 2 of 4, so (4)(3)/(2) = 6 possibilities. For the first side dish, we have 4 options, but for the second side dish we have to choose from the 3 remaining options. We divide by 2 because the order in which the two side dishes are selected doesn’t matter. For example, picking mashed potatoes then side salad is the same as picking side salad then mashed potatoes.
Drink selection: choose 1 of 7, so 7 possibilities.
Thus, there are (3)(6)(7) = 126 different combos of the first type.
The second type of combo is 1 soup, 3 different side dishes, and 1 drink. When a customer orders this type, she makes the following decisions:
Soup selection: choose 1 of 3, so 3 possibilities.
Side dish selection: choose 3 of 4, which is like choosing which 1 of the 4 NOT to order, so 4 possibilities.
Drink selection: choose 1 of 7, so 7 possibilities.
The total number of this type of combination is thus (3)(4)(7) = 84 combos.
Put together, the customer could order either the first type of combo OR the second type of combo, so she had 126 + 84 = 210 unique lunch options. Note that “and” means multiply on combo problems, which is why we multiplied when the customer could order food “and” a drink. “Or” means add on combo problems, which is why we added when the customer could order either the entree lunch combo “or” the soup lunch combo.
The correct answer is D.
MGRE's explanation is about the same as mine, and we both got to the same answer by the same means. In the end, you have to calculate the number of possible configurations for each combo, then add those two sets of possibilities together for your total.
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