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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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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\(6P_4 = 360\)

Answer must be (E) 360

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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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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.

Answer: E
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total choices ; 6*5*4*3 ; 360
IMO E

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