Bunuel wrote:
Patronus wrote:
A 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) \((6*10*4) + (6*\frac{(10*9)}{2} * 4)\)
B) 6*10*4
C) 6*10*2*4
D) 6*9*4
E) \(\frac{6*10*4}{2}\)
1 appetizer out of 6, 1 entrees out of 10, 1 dessert out of 4: \(6*10*4\);
1 appetizer out of 6, 2 entrees out of 10, 1 dessert out of 4: \(6*C^2_{10}*4=6*\frac{(10*9)}{2} * 4\).
Total = A.
Answer: A.
Thank you for the short answer Bunuel.
Can you please tell me why we should use \(\frac{(10*9)}{2}\) and not 10*9?
I solved it by thinking we have to fill 2 spots out of 10 contenders (entrees). Therefore, first spot can be filled in 10 ways and the 2nd spot in 9 ways.
Sorry, but I normally get confused with when to use Combinations and Permutations. Can you please clarify?