Question 1120025
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Since it is possible to not have an appetizer, there are actually 7 choices for the appetizer course.  Similarly, there are 9 entrée choices counting not having the main course at all, and there are 6 dessert choices.  So for each of the 7 appetizers, you have 9 entrée choices for a total of 63 choices for the appetizer and entrée.  Then for each of those 63 choices, there are 6 dessert choices, 63 times 6 is 378 possibilities.  However, this answer is one too large because it includes the possibility of having nothing at all which is not allowed by the condition "A customer can choose to eat just one course, or two different courses, or all three courses."  Therefore the final answer is 377 different possible meals.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}

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