GRE Sample Test 4 Quantitative Reasoning 1

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 $\underset{\mathit{n}}{}{\mathit{C}}_{\mathit{k}}$, is given by the formula
$\underset{\mathit{n}}{}{\mathit{C}}_{\mathit{k}}$ = $\frac{\mathit{n}!}{\mathit{k}!\left(\mathit{n}-\mathit{k}\right)!}$
Here the total number of different possible servings for a plate is $\left(\underset{5}{}{\mathit{C}}_1\right)\left(\underset{5}{}{\mathit{C}}_2\right)$ + $\left(\underset{5}{}{\mathit{C}}_2\right)\left(\underset{4}{}{\mathit{C}}_1\right)$ + $\left(\underset{5}{}{\mathit{C}}_3\right)$
Now $\left(\underset{5}{}{\mathit{C}}_3\right)$ represents choosing 1 type of vegetable selection from 5 different types, so $\left(\underset{5}{}{\mathit{C}}_3\right)$ = 5. (The formula also gives this result.) Now we use the formula to find the next two combinations:
$\left(\underset{4}{}{\mathit{C}}_2\right)$ = $\frac{4!}{2!\left(4-2\right)!}$ = $\frac{4!}{2!.2!}$ = $\frac{4.3.2.1}{2.1.2.1}$ = 6
$\left(\underset{5}{}{\mathit{C}}_2\right)$ = $\frac{5!}{2!\left(5-2\right)!}$ = $\frac{5!}{2!.3!}$ = $\frac{5.4.3.2.1}{2.1.3.2.1}$ = 10
Here $\left(\underset{4}{}{\mathit{C}}_1\right)$ corresponds to choosing 1 type of meat selection from 4 different types, so $\left(\underset{4}{}{\mathit{C}}_1\right)$ = 4. Then we use the formula again:
$\left(\underset{5}{}{\mathit{C}}_3\right)$ = $\frac{5!}{3!\left(5-3\right)!}$ = $\frac{5!}{3!.2!}$ = $\frac{5.4.3.2.1}{3.2.1.2.1}$ = 10
So the number of different possible dinners of these three items is 5 × 6 + 10 × 4 + 10 = 80, choice (E).