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?

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)

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...

To determine the possibilities, we need to calculate the maximum number of different choices that can be made in each scenario.

Scenario 1: Each passenger took one snack, one dinner, and one beverage.

For the snacks, there are two options (peanuts or pretzels). For the dinner, there are also two options (chicken or beef). And for the beverages, there are 12 options. To calculate the total number of different choices, we multiply these numbers together:

2 (snack options) * 2 (dinner options) * 12 (beverage options) = 48 different choices

However, there are 138 passengers on the flight, and the flight attendant remarked that no two passengers had the same request. Since the number of different choices (48) is less than the number of passengers (138), it is not possible for each passenger to have a unique selection in this scenario.

Scenario 2: Each passenger took no more than 1 selection from each category.

In this scenario, each passenger can choose one snack (peanuts or pretzels), one dinner (chicken or beef), and one beverage. The total number of different choices can be determined using the same approach as before:

2 (snack options) * 2 (dinner options) * 12 (beverage options) = 48 different choices

Again, since the number of different choices (48) is less than the number of passengers (138), it is not possible for each passenger to have a unique selection in this scenario.

Scenario 3: Each passenger took no more than one dinner or beverage, and the snack options were "peanuts," "pretzels," or "both."

In this scenario, each passenger can choose either peanuts, pretzels, or both for the snack. For the dinner, they can choose chicken or beef (only one option allowed), and for the beverage, they can choose from 12 options (only one option allowed).

If we calculate the total number of different choices:
3 (snack options) * 2 (dinner options) * 12 (beverage options) = 72 different choices

Since the number of different choices (72) is greater than the number of passengers (138), it is indeed possible for each passenger to have a unique selection in this scenario.