gmat_winter wrote:Szechuan House offers chicken, shrimp, beef, vegetable, and tofu entrees served in either garlic sauce or green chili sauce. In how many ways can three entrees be served if no entree is served more than a total of once?
A)10
B)60
c)80
D)100
E)120
OAC
Solution:
This is a combination question testing us on the fundamental counting principle. The first step is to determine the number of ways in which we can select 3 dishes. We are selecting 3 dishes from 5 so we have 5C3, which is calculated as follows:
5C3 = 5!/[(5-3)! x 3!] = 5!/2! x 3! = (5x4x3x2x1)/(2x1x3x2x1) = 120/12 = 10
If you don't know how to use the combination formula, here is a method that will work equally well.
We are choosing 3 entrées from a pool of 5 possible entrées. That is, there are 3 decisions being made:
Decision 1: Choosing the first entrée
Decision 2: Choosing the second entrée
Decision 3: Choosing the third entrée
Five different entrées are available to be the first selection.
For the second selection, 4 remaining entrées are available because 1 entrée was already chosen.
For the third selection, 3 remaining entrées are available remain because 2 entrées have already been selected.
The product of these options will represent the numerator of our calculation, so we will have: 5 x 4 x 3 = 60
The final step is to divide by the factorial of the number of "decisions." In this case, we had 3 decisions so we divide by divide by 3!.
(5x4x3)/3! = (5x4x3)/(3x2x1) = 60/6 = 10
We see that there are 10 ways to choose 3 entrées from 5 available entrées. That is, there are ten 3-entrée combos. For each entrée in any of these 3-entrée combos, there are two selections of the sauces. That is, if entrée one, entrée two and entrée three are in the 3-entrée combo, we can serve each with either garlic sauce or green chili sauce. Therefore, for each 3-entrée combo, there are a total of 2 x 2 x 2 = 8 ways to serve them with a sauce.
If this is difficult to see, let's say a 3-entrée combo consists of chicken (C), shrimp (S) and beef (B) and let's let g be garlic sauce and c be green chili sauce. We will have:
1) C-g, S-g, B-g
2) C-g, S-g, B-c
3) C-g, S-c, B-g
4) C-g, S-c, B-c
5) C-c, S-g, B-g
6) C-c, S-g, B-c
7) C-c, S-c, B-g
8) C-c, S-c, B-c
Finally, since there are ten 3-entrée combos and each combo has 8 selections, there are a total of 10 x 8 = 80 ways to pick three entrées (each with a sauce) where no entrée is served more than once.
The answer
C
