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# A family of 4 is at a restaurant with 6 choices on the menu

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Joined: 21 Jun 2014
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A family of 4 is at a restaurant with 6 choices on the menu [#permalink]

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10 Mar 2017, 22:50
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Difficulty:

35% (medium)

Question Stats:

54% (00:38) correct 46% (00:38) wrong based on 85 sessions

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A family of 4 is at a restaurant with 6 choices on the menu. If nobody orders the same meal, how many possible combinations of meals could they order?

A) 15

B) 30

C) 60

D) 180

E) 360
[Reveal] Spoiler: OA

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Re: A family of 4 is at a restaurant with 6 choices on the menu [#permalink]

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10 Mar 2017, 22:59
2
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HKD1710 wrote:
A family of 4 is at a restaurant with 6 choices on the menu. If nobody orders the same meal, how many possible combinations of meals could they order?

A) 15

B) 30

C) 60

D) 180

E) 360

Person 1 can order from 6 different choices
Person 2 can order from 5 different choices
Person 3 can order from 4 different choices
Person 4 can order from 3 different choices

total choices = 6*5*4*3 =360

Hence option E is correct
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Intern
Joined: 22 Dec 2013
Posts: 9
Location: United States
Schools: Wharton '20
Re: A family of 4 is at a restaurant with 6 choices on the menu [#permalink]

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10 Mar 2017, 23:09
HKD1710 wrote:
A family of 4 is at a restaurant with 6 choices on the menu. If nobody orders the same meal, how many possible combinations of meals could they order?

A) 15

B) 30

C) 60

D) 180

E) 360

Person 1 has all the 6 items to choose from so he can choose in 6 ways
Now since nobody orders the same meal and since person 1 has already selected his choice of meal, person 2 has 5 ways to choose
similarly person 3 has 4 ways and person 4 has 3 ways
Total ways = 6*5*4*3 =360

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Re: A family of 4 is at a restaurant with 6 choices on the menu [#permalink]

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11 Mar 2017, 07:14
HKD1710 wrote:
A family of 4 is at a restaurant with 6 choices on the menu. If nobody orders the same meal, how many possible combinations of meals could they order?

A) 15

B) 30

C) 60

D) 180

E) 360

$$6P_4 = 360$$

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Re: A family of 4 is at a restaurant with 6 choices on the menu [#permalink]

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12 Mar 2017, 09:28
Hi,

Can Bunuel or VeritasPrepKarishma help me.

Can you please explain to me how comes you use Permutations in this case?
I know we use Permutations when the order matters. While, we use Combinations when the order doesn't matter.
Can you elaborate how the order matters in this case.
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Re: A family of 4 is at a restaurant with 6 choices on the menu [#permalink]

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14 Mar 2017, 00:18
1
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Expert's post
matthewsmith_89 wrote:
Hi,

Can Bunuel or VeritasPrepKarishma help me.

Can you please explain to me how comes you use Permutations in this case?
I know we use Permutations when the order matters. While, we use Combinations when the order doesn't matter.
Can you elaborate how the order matters in this case.

Question: how many possible combinations of meals could they order?
which means "In how many ways can they order?"

There are 4 people: say A, B, C, D
There are 6 meals: say M1, M2, ... M6

You have 6 meals and you have to choose 4 of them - this is a combination problem
Now, you have to distribute the 4 meals among the 4 people - this is permutation
A - M1, B - M2, C - M3, D - M4
is different from
A - M4, B - M2, C - M3, D - M1

A ordering M1 is different from A ordering M4 and hence the 4 meals need to be arranged among A, B, C and D.

Had the question been: A person arrives at a restaurant to pick 4 meals to take home. He notices that the menu has 6 meal options. In how many ways can he pick the 4 meals?

Here, this is just selection - a combination problem.
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Re: A family of 4 is at a restaurant with 6 choices on the menu [#permalink]

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15 Mar 2017, 15:53
1
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Expert's post
HKD1710 wrote:
A family of 4 is at a restaurant with 6 choices on the menu. If nobody orders the same meal, how many possible combinations of meals could they order?

A) 15

B) 30

C) 60

D) 180

E) 360

Since each choice on the menu is unique and each person has to order a different meal, order DOES MATTER, and thus we have a permutation problem.

The number of ways for a family of 4 to select from 6 different choices on menu, if each person must select a different meal, is 6P4 = 6!/(6-4)! = 6 x 5 x 4 x 3 = 360 ways.

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Re: A family of 4 is at a restaurant with 6 choices on the menu   [#permalink] 15 Mar 2017, 15:53
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