This Week's Problem: "Lunch Combos"
Q:
The standard lunch price at a cafeteria buys either a combination of 1 entree, 2 different side dishes, and 1 drink or a combination of 1 soup, 3 different side dishes, and 1 drink. Substitutions are not allowed, and customers cannot order multiple servings of any one side dish. If customers can choose from among 3 entree, 4 side dish, 3 soup, and 7 drink options, how many different lunch combinations are available for the standard lunch price?
A:
(A) 54
(B) 84
(C) 126
(D) 210
(E) 252
Go for it. My answer will appear in the comments, and MGRE will publish the official answer next week. My first impression is that we're dealing with a typical permutations/combinations problem... but we'll see. I need to stare at it for a moment.
_
3 comments:
I'm not very good with permutations and combinations, but I say it's D. The basic method is this: to find the number of possible choices for each respective combo-- we'll call them A and B-- you need to multiply:
For Combo A
[# possible entrees] * [# possible side dish combinations] * [# possible drinks]
For Combo B
[# possible soups] * [# possible side dish combinations] * [# possible drinks]
This becomes, respectively, 3*6*7 and 3*4*7. That's 126 and 84, which together add up to 210.
Another reason why I think the answer is D is that, even before we do all that multiplication, we see very quickly that Combos A and B are, respectively, multiples of 21. Adding two multiples of 21 will give us another multiple of 21, which is just what 210 is. Answer A doesn't fit the bill, and B and C are obviously too small, which leaves us with only D and E as plausible choices even before we do any serious math.
Your thoughts?
So how did I figure those side dish combinations? Well, for Combo B, it was easy: you're allowed to choose 3 out of 4 possible sides, which means that, no matter which sides you choose, there's only one side dish left unchosen. If each side "takes a turn" at being unchosen, then the restaurant's patron has only 4 possible choices. To spell it out:
Sides 1,2,3 (4 left unchosen)
Sides 1,2,4 (3 left unchosen)
Sides 1,3,4 (2 left unchosen)
Sides 2,3,4 (1 left unchosen)
Easy enough. For Combo A, however, things are a bit more complicated, but if you don't know the appropriate formula to use here, you can just do a plug-and-play, as I did. Combo A allows only 2 sides out of a possible 4, so that comes out like this:
Sides 1,2
Sides 1,3
Sides 1,4
Sides 2,3
Sides 2,4
Sides 3,4
Only 6 possibilities to cover the gamut of choices.
According to MGRE, my answer is correct! 210 it is.
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