A restaurant menu features 5 appetizers, 6 entrées, and 3 desserts. If

Question

A restaurant menu features 5 appetizers, 6 entrées, and 3 desserts. If a dinner special consists of 1 appetizer, 1 entrée, and 1 dessert, how many different dinner specials are possible?

A. 30
B. 60
C. 90
D. 120
E. 150

Lampard42 · Answer

1 appetizer, 1 entrée, and 1 dessert = 5 options* 6 options* 3 options = 90 IMO C

CareerGeek · Answer

Total number of ways = 5c1*6c1*3c1 = 90

Option C

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Kinshook · Answer

Number of different dinners = 5C1 * 6C1 * 3C1 = 5*6*3 = 90

IMO C

rpkr · Answer

Five ways to select appetizer,six ways to select entrée, and 3 ways to select dessert, 
Total: 5*6*3 = 90

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Archit3110 · Answer

total options
5c1*6c1*3c1; 90
IMO C

CrackverbalGMAT · Answer

This is a very easy question on Combinations. It’s also an apt question for anyone who’s starting off with the topic and wants to build a strong foundation, because solving such simple questions will give you a good idea of the fundamental concepts.

The number of ways of selecting ANY r objects out of the available n objects =  n_c_r  =  \frac{n!}{(n-r)! * r!}

Since the dinner special has to contain ANY 1 appetizer, ANY 1 entree and ANY 1 dessert, we need to select the above from the available 5 appetizers, 6 entrees and 3 desserts.

No. of ways of selecting 1 appetizer out of 5 appetizers =  5_C_1  = 5. { n_c_1  = n}

No. of ways of selecting 1 entrée out of 6 entrees =  6_C_1  = 6.

No. of ways of selecting 1 dessert out of 3 desserts =  3_C_1  = 3.

Therefore, number of ways of selecting 1 appetizer, 1 entrée AND 1 dessert = 5 * 6 * 3 = 90. 
The correct answer option is C.

Hope this helps!

ScottTargetTestPrep · Answer

The number of possible dinner specials is 5 x 6 x 3 = 90.

Answer: C

bumpbot · Answer

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A restaurant menu features 5 appetizers, 6 entrées, and 3 desserts. If

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