MHB How many possible lunch special combinations can be ordered

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SUMMARY

The discussion centers on calculating the total number of lunch special combinations at Deb's Deli, where customers can choose between a sandwich with a salad or a sandwich with a soup. There are 5 sandwich options, 4 salad options, and 3 soup options available. The correct approach to find the total combinations is to apply the fundamental counting principle, resulting in the formula N = 5(4 + 3), which simplifies to N = 5 x 7 = 35 possible combinations.

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Hi All, I'm studying for the GRE, and really struggling with combination questions for some reason. I'm posting here quite a bit, but just wanted to say thank you so much for your help.

What would be the fastest way to solve the following?

"At Deb's Deli, a customer may choose either a sandwich and a salad or a sandwich and a soup for the lunch special. There are 5 choices of sandwich, 4 choices of salad, and 3 choices of soup. How many possible lunch special combinations can be ordered"
 
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greprep said:
Hi All, I'm studying for the GRE, and really struggling with combination questions for some reason. I'm posting here quite a bit, but just wanted to say thank you so much for your help.

What would be the fastest way to solve the following?

"At Deb's Deli, a customer may choose either a sandwich and a salad or a sandwich and a soup for the lunch special. There are 5 choices of sandwich, 4 choices of salad, and 3 choices of soup. How many possible lunch special combinations can be ordered"

One way would be to compute the number of possible sandwich/salad combinations are possible, then compute the number of sandwich/soup combinations are possible, and then add the two to get the total. Or, you could do it in one pass by looking at it as a sandwich/item problem, where you have 5 choices for sandwich and 7 choices for item (salad or soup) and apply the fundamental counting principle.

What do you get?
 
Would it then just be 5x4x3, according to the fundamental counting principal?
 
greprep said:
Would it then just be 5x4x3, according to the fundamental counting principal?

No, not quite...you have 5 sandwich options, and then for the second option, you have 3 + 4 options:

$$N=5(4+3)=\,?$$
 

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