Help on combination problem

[crackgmat007]

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

OA:E

I tried solving as 6C4 - 15. Am I missing something? Let me know. Thanks.

[piyush_nitt]

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

OA:E

I tried solving as 6C4 - 15. Am I missing something? Let me know. Thanks.

Hey

The reason why you got a wrong answer because you have ignored the arrangement. Instead of 6C4 it should be 6P4.

[Vemuri]

Well, it would be less confusing to just apply logic:

If the 1st person were to order a meal, he/she has 6 choices.
The second person will be left with only 5 choices (because same meal cannot be ordered)
The third person will be left with only 4 choices
The fourth person will be left with only 3 choices.

Thus, the number of possible combinations is = 6*5*4*3 = 360

Hope this helps.

[crackgmat007]

thanks..i got the logic

[lunarpower]

this is a problem in which "order matters" -- i.e., if you scramble anything in your selection, then the result constitutes an entirely new possibility.

in problems on which "order matters", the easiest method by far is plain old consecutive multiplication. in other words, just write the number of possibilities for each choice, and then multiply them together.
in this case, there are 6 choices for the first selection. then, because no repetitions are allowed, there are only 5 choices for the next selection, 4 for the third, and 3 for the fourth.

therefore, the total number of possibilities is 6 x 5 x 4 x 3, or 360.

--

in general, using "permutation formulas" (such as the "6p4" mentioned in one of the previous posts) is essentially a waste of time, since such formulas are merely indirect ways of writing the same product you'd get with consecutive multiplication. in other words, if you actually wrote 6p4 = 6!/2! on this problem, you'd have to waste a non-negligible amount of time writing those factorials out and cancelling - only to leave 6x5x4x3, the same product you could have written directly in the first place.

here's the upshot, then:
combination formulas can be useful.
permutation formulas, not so much. you should just leave these alone and stick witch consecutive multiplication.

[tanviet]

I do not understand. I thing correct answer is A.

think that we fill 6 choice in 4 slot as princeton review did. so, we have 6x5x4x3 choice.

but in this, ORDER DOSE NOT MATTER, because whether the family order ABCD or BADC is the same.

so 6x5x4x3/4! =15

pls, help if I make mistake

[lunarpower]

I do not understand. I thing correct answer is A.

think that we fill 6 choice in 4 slot as princeton review did. so, we have 6x5x4x3 choice.

but in this, ORDER DOSE NOT MATTER, because whether the family order ABCD or BADC is the same.

so 6x5x4x3/4! =15

pls, help if I make mistake

ah, ok, i see what you mean.

in fact, the wording of the problem is ambiguous. it's not clear which of the following is the case:
* each member of the family orders a plate individually (in which case order matters); or
* the family orders four plates communally (in which case order doesn't matter).


i assumed that the former is the case, primarily because of the wording "nobody orders the same meal". i took those words to mean that each member of the family individually ordered a plate, so that order matters (i.e., if the same 4 plates are ordered, but by different people, then that's a different outcome).

i can see, though, how you would have interpreted it the other way.

--

rest assured that official problems will not have this degree of ambiguity in their problem statements.

[Jeff@TargetTestPrep]

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.

Answer: E