The lunch menu at a certain restaurant contains 4 different entrees an

Take the task of creating a meal and break it into stages.

Stage 1: Select an entree
There are four different entrees from which to choose.
So, we can complete stage 1 in 4 ways

Stage 2: Select 2 different side dishes
Since the order in which we select the 2 side dishes does not matter, we can use combinations.
We can select 2 side dishes from 5 side dishes in 5C2 ways = (5)(4)/(2)(1) = 10
So we can complete stage 2 in 10 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a meal) in (4)(10) ways (= 40 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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For the first side dish there 5 possible options, for the second only 4 (since we are looking for different combinations)
As it says "and", we have to multiply both combinations: 4 x (5x4) = 80 different combinations

total options ;
4c1*5c2 ; 4*10 ; 40
IMO C

\(\binom{4}{1}\binom{5}{2}=4\cdot\frac{5!}{2!\cdot3!}=4\cdot10=40\)

Answer: C

The question is very clear that we want different combinations and not arrangements.


Suppose the entry dishes are E1, E2, E3 and E4

Let the side dishes be S1, S2, S3, S4 and S5

If we take the side dishes 2 at a time, the different possible combinations are S1S2, S1S3, S1S4, S1S5, S2S3, S2S4, S2S5, S3S4, S3S5, S4S5 = 10 combinations. Order does not matter here, and therefore S1S2 = S2S1

Now with E1, I can have 10 options of side dishes. Similar is the case with each of the other 3 entrees.

Therefore total combinations = 10 * 4 = 40


Option C

Arun Kumar

The number of combinations is 4C1 x 5C2 = 4 x (5 x 4)/2! = 4 x 10 = 40

Answer: C
_________________

1 of 4 different entrees can be selected in 4 ways
2 of 5 different side dishes can be selected in 5C2 = 10 ways

Thus, total no. of possible combinations: 4x10 = 40 ways (Option C)