Jane must select three different items for each dinner she will serve. The items are to be chosen from among five different vegetarian and four different meat selections. If at least one of the selections must be vegetarian, how many different dinners could Jane create?

A. 30

B. 40

C. 60

D. 70

E. 80

E. 80

Step-by-step explanation:

First consider exactly what the problem is asking. You have to make a dinner and AT LEAST one of the dishes has to be vegetarian. So then we could have all three be vegetarian (V V V), 2 vegetarian with one meat (V V M), or one vegetarian with 2 meat (V M M). If I can find the number of arrangements possible for each of the possibilities and add them, we should have our answer.

Consider first the all vegetarian option. There are 5 vegetarian meals on the menu and we may choose 3 of them. This will be a combination since the order in which we choose doesn't matter. For example if my 5 vegetarian dishes are salad, hummus, rice, lentils and pasta, it doesn't matter what order I serve them in, since they will all be a part of the meal. Simply put, placing hummus, rice, and lentils on the table is the same as placing rice, lentils and hummus on the table if everybody shares the dishes. If I had specific guests assigned to the each dish then the order would matter, and it would be a permutation. For example if three of Jane's guests, (lets say Mike, Frank, and Bob) are going to have a specific dish, then the arrangement where Mike has rice, Bob has lentils, and Frank has pasta is different from the arrangement where Mike has lentils, Bob has rice, and Frank has pasta. Since there is no mention of a specific order this has to go in, it is safe to assume a combination. So how many ways can we choose 3 vegetarian dishes from 5 options? This will be 5 C 3.

So we found that (V V V) gives us 5 C 3, so lets examine the other remaining cases. (V V M) implies from 5 vegetarian options we can choose 2 and from 4 meat options we choose one. Then (V V M) gives us 5 C 2 * 4 C 1. Likewise for (V M M) we can say 5 C 1 * 4 C 2.

Putting it all together we have 5 C 3 + 5 C 2 * 4 C 1 + 5 C 1 * 4 C 2. 10 + 10*4 + 5 * 6 = 80.