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

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Joined: 21 Aug 2014
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GMAT 1: 610 Q49 V25
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A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]

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24 Jun 2015, 09:42
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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) $$\frac{6*10*4}{2}$$
[Reveal] Spoiler: OA

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Joined: 02 Sep 2009
Posts: 43864
Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]

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24 Jun 2015, 09:50
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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.

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Joined: 21 Aug 2014
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GMAT 1: 610 Q49 V25
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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]

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24 Jun 2015, 09:57
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.

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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Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]

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24 Jun 2015, 10:03
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Expert's post
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.

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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A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]

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25 Jun 2015, 00:39
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: $$6*C^2_{10}*4=6*\frac{(10*9)}{2} * 4$$.

Hi Bunnel,

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

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 43864
Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]

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25 Jun 2015, 01:15
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: $$6*C^2_{10}*4=6*\frac{(10*9)}{2} * 4$$.

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.

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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Posts: 790
Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an [#permalink]

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02 Sep 2017, 13:48
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
Re: A restaurant serves 6 varieties of appetizers, 10 different entrees an   [#permalink] 02 Sep 2017, 13:48
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