The question asks for the number of different dinners Jane could make. Since the order of the selections in the dinner doesn’t matter, this is a combination problem. But it involves three possible combination types: Veg, Meat, Meat; Veg, Veg, Meat; or Veg, Veg, Veg. We must calculate the possibilities for each type of combination and then add the results to find the total number of different combinations possible.
Let V represent vegetarian and M represent meat.
Then with V, M, M, she has 5 choices for the vegetarian (she must choose 1) × 4 choices for meat (she must choose 2).
For V, V, M, she will choose 2 from among 5 for the vegetarian and 1 among 4 for the meat.
If she goes with V, V, V, the all-vegetarian menu, she will choose a subgroup of 3 from among 5 vegetarian choices.
If n and k are positive integers where n = k, then the number of different subgroups consisting of k objects that can be selected from a group consisting of n different objects, denoted by
nCk, is given by the formula
nCk =
k!(n−k)!n! Here the total number of different possible servings for a plate is
(5C1)(5C2) +
((5C2)(4C1)) +
(5C3) Now
(5C3) represents choosing 1 type of vegetable selection from 5 different types, so
(5C3) = 5. (The formula also gives this result.) Now we use the formula to find the next two combinations:
(4C2) =
2!(4−2)!4! =
2!⋅2!4! =
2⋅1⋅2⋅14⋅3⋅2⋅1 = 6
(5C2) =
2!(5−2)!5! =
2!⋅3!5! =
2⋅1⋅3⋅2⋅15⋅4⋅3⋅2⋅1 = 10
Here
(4C1) corresponds to choosing 1 type of meat selection from 4 different types, so
(4C1) = 4. Then we use the formula again:
(5C3) =
3!(5−3)!5! =
3!⋅2!5! =
3⋅2⋅1⋅2⋅15⋅4⋅3⋅2⋅1 = 10
So the number of different possible dinners of these three items is 5 × 6 + 10 × 4 + 10 = 80, choice (E).