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A restaurant serves 6 varieties of appetizers, 10 different entrees an

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A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   24 Jun 2015, 10:42
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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) \(\frac{6*10*4}{2}\)
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Bunuel
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   24 Jun 2015, 10:50
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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.
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   24 Jun 2015, 11:03
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Patronus wrote:
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?


\(C^2_{10}\), which is 10*9/2 is the number of two entrees possible when the order of entrees does not matter.

\(P^2_{10}\), which is 10*9 is the number of two entrees possible when the order of entrees matters, i.e. when we differentiate between (entree X, entrees Y) from (entree Y, entrees X).
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   24 Jun 2015, 10:57
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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?
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A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   25 Jun 2015, 01:39
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: \(6*C^2_{10}*4=6*\frac{(10*9)}{2} * 4\).


Answer: A.


Hi Bunnel,

Why is there (+) in choice A? what does it mean? what not multiplication sign (*)?

Thanks
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   25 Jun 2015, 02:15
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Mo2men wrote:
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: \(6*C^2_{10}*4=6*\frac{(10*9)}{2} * 4\).


Answer: A.


Hi Bunnel,

Why is there (+) in choice A? what does it mean? what not multiplication sign (*)?

Thanks


Principle of Multiplication
If an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘m*n’ ways.

Principle of Addition
If an operation can be performed in ‘m’ different ways and another operation in ‘n’ different ways then either of these two operations can be performed in ‘m+n’ ways (provided only one has to be done).
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   02 Sep 2017, 14:48
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) \(\frac{6*10*4}{2}\)


There's two basic scenarios

6c1 x 10c1 x 4c1 = [6 x 10 x 4]

6c1 x 10c2 x 4c1 = [6 x 45 x 4]

Account for both possibilities hence
How many possibilities with 10c1 + how many possibilities with 10c2
[6 x 10 x 4] + [6 x 45 x 4]

A
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink] New post   11 Sep 2022, 06:40
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]
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