Homework Help: College math

Posted by Amy on Thursday, December 2, 2010 at 2:09pm.

On a flight with 138 passengers, each passenger had the choice of peanuts or pretzels for a snack and the coice of chicken or beef for dinner. The beverage service offered 12 beverages. A flight attendant later remarked that no 2 passengers had the same request. Is this possible if each passenger took one snack, one dinner, and one beverage? Is this possible if each passenger took no more than 1 selection from each category? Is this possible if the flight attendant, being a new recruit, offered the snack by asking, "peanuts, pretzels, or both?" and each passenger took no more than one dinner or beverage?

College math - Ann Hiro, Thursday, December 2, 2010 at 11:49pm
The first two questions seem the exact same, oddly; perhaps I'm misunderstanding, but going on the assumption that I'm not...

There are 48 different combinations of grub (12 x 2 x 2), so the first two are 'no'.

When the passengers could elect to take both snacks, this becomes 72 combinations (12 x 3 x 2), again 'no'.

Even if the passengers could elect to not take a snack, meal, or beverage, there still would not be enough combinations for everyone to have a unique order (13 x 3 x 3 = 117, < 138)
College math - Ann Hiro, Thursday, December 2, 2010 at 11:51pm
I knew I was missing something...

On the last bit, if passengers could take both of the snacks, AND elect to take nothing from any of the three categories, you get 156 combinations (13 x 4 x 3), so THAT would indeed work...

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