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April 25, 2014

April 25, 2014

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

- College math -
**Ann Hiro**, Thursday, December 2, 2010 at 11:49pmThe 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 11:51pmI 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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