Question: The restaurant serves 6 varieties of appetizers, 10 different entrees and 4 different desserts. In how many ways can one make a meal if one chooses an appetizer, at least one and at most two different entrees and one dessert?

A) (6 ∗ 10 ∗ 4) + (6 ∗ (10∗9)/2 ∗ 4)
B) 6*10*4
C) 6*10*2*4
D) 6*9*4
E) (6∗10∗4)/2

Correct Answer: A
Solution and Explanation:
Approach Solution 1:

The restaurant has total varieties of appetizers = 6;
Total no of entrees = 10
Total no of desserts = 4
The question asks to find the number of ways one can make a meal if one chooses an appetizer, at least one and at most two different entrees and one dessert.

Therefore, the required number of ways = 1 appetizer + at least one and at most two different entrees + one deserts.
1 appetizer out of 6 is 6C1
1 entree out of 10 is 10C1
1 dessert out of 4 is 4C1
The total number of possible ways from the above is 6C1 x 10C1 x 4C1 = 6 ∗ 10 ∗ 4.

1 appetizer out of 6 is 6C1
1 entree out of 10 is 10C2
1 dessert out of 4 is 4C1

The total number of possible ways from the above is 6C1 x 10C2 x 4C1= 6 ∗ 45 ∗ 4
The required no of ways for the above-given conditions is the sum of the multiplication of the two ways:
6 ∗ 10 ∗ 4 + 6 ∗ 45 ∗ 4
Or we can rewrite it as (6 ∗ 10 ∗ 4) + (6 ∗ (10∗9)/2 ∗ 4).

Approach Solution 2:

The problem statement states that:

Given:

• The restaurant serves 6 varieties of appetizers, 10 different entrees and 4 different desserts.

The question asks to find the number of ways one can make a meal if one chooses an appetizer, at least one and at most two different entrees and one dessert.

The number of ways of selecting any objects out of the available n objects = n_c_r = n! / (n−r)!∗r!
Since the dinner special must contain ANY 1 appetizer, ANY 1 entree, and ANY 1 dessert, we need to select the above from the available 6 appetizers, 10 entrees and 4 desserts.

No. of ways of selecting 1 appetizer out of 6 appetizers = 6_C_1 = 6. {n_c_1 = n}
No. of ways of selecting 1 entrée out of 10 entrees = 10_C_1 = 10.
No. of ways of selecting 1 dessert out of 3 desserts = 4_C_1 = 4.
Therefore, the number of ways of selecting 1 appetizer, 1 entrée, and 1 dessert = 10* 6 * 4.

No. of ways of selecting 1 appetizer out of 6 appetizers = 6_C_1 = 6. {n_c_1 = n}
No. of ways of selecting 2 entrée out of 110 entrees = 10_C_2 = 10*9/2.
No. of ways of selecting 1 dessert out of 3 desserts = 4_C_1 = 4.
Therefore, the number of ways of selecting 1 appetizer, 1 entrée, and 1 dessert = 10*9* 6 * 4/2.
The total required ways are 6 ∗ 10 ∗ 4 + 6 ∗ (10*9)/2 ∗ 4.

“The restaurant serves 6 varieties of appetizers, 10 different entrees”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. GMAT Problem Solving questions stimulate the candidates to assess every condition stated in the question to solve numerical problems. GMAT Quant practice papers help the candidates to get familiar with different sorts of questions that will improve their mathematical learning.

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