An airline serving n cities has 2 nonstop flights per day, one in each direction, between each pair of these...

1. Translate the problem requirements: We need to clarify what "2 nonstop flights per day, one in each direction, between each pair of cities" means - this refers to bidirectional flights between every possible pair of cities among n total cities.
2. Count the number of city pairs: Determine how many unique pairs of cities can be formed from n cities, since each pair will have flights between them.
3. Account for bidirectional flights: Since there are 2 flights per day between each pair (one in each direction), multiply the number of pairs by 2 to get total flights.
4. Verify with concrete example: Test the formula with a small number of cities to ensure our logic is correct.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this airline situation means in everyday terms. We have n cities, and the airline wants to connect every possible pair of cities with flights. When the problem says "2 nonstop flights per day, one in each direction, between each pair of these cities," it means:

• If you can fly from City A to City B, you can also fly from City B to City A
• These count as 2 separate flights (one going each way)
• This happens between EVERY possible pair of cities

For example, if we have cities New York, Boston, and Chicago, then we need flights:

• New York \(\leftrightarrow\) Boston (2 flights: NY\(\rightarrow\)Boston and Boston\(\rightarrow\)NY)
• New York \(\leftrightarrow\) Chicago (2 flights: NY\(\rightarrow\)Chicago and Chicago\(\rightarrow\)NY)
• Boston \(\leftrightarrow\) Chicago (2 flights: Boston\(\rightarrow\)Chicago and Chicago\(\rightarrow\)Boston)

Process Skill: TRANSLATE - Converting the problem language into mathematical understanding

2. Count the number of city pairs

Now we need to figure out how many unique pairs of cities we can make from n cities total.

Let's think about this step by step:

• The first city can be paired with (n-1) other cities
• The second city can be paired with (n-2) remaining cities (we don't double-count the pair with the first city)
• The third city can be paired with (n-3) remaining cities
• And so on...

This gives us: (n-1) + (n-2) + (n-3) + ... + 1

But there's a simpler way to think about it. If we have n cities, we're choosing 2 cities at a time to form pairs. The number of ways to choose 2 items from n items is \(\mathrm{n(n-1)/2}\).

Let's verify with our 3-city example:

• Number of pairs = \(\mathrm{3(3-1)/2 = 3(2)/2 = 3}\) pairs
• We identified: NY-Boston, NY-Chicago, Boston-Chicago = 3 pairs ✓

3. Account for bidirectional flights

Now we know there are \(\mathrm{n(n-1)/2}\) unique pairs of cities. But remember, between each pair, there are 2 flights (one in each direction).

Total number of flights = (Number of city pairs) × (2 flights per pair)
Total number of flights = \(\mathrm{n(n-1)/2 \times 2 = n(n-1)}\)

Let's check with our 3-city example:

• Total flights = \(\mathrm{3(3-1) = 3(2) = 6}\) flights
• We counted: NY\(\rightarrow\)Boston, Boston\(\rightarrow\)NY, NY\(\rightarrow\)Chicago, Chicago\(\rightarrow\)NY, Boston\(\rightarrow\)Chicago, Chicago\(\rightarrow\)Boston = 6 flights ✓

The answer is B.