To solve this problem, we need to calculate how many different ways the firm can select 6 cities from a total of 13 possible cities. This is a combination problem, as the order of selection does not matter.

The formula for combinations is given by:

$C(n, k) = \frac{n!}{k!(n-k)!}$

where:

• $n$ is the total number of items (in this case, 13 cities),
• $k$ is the number of items to choose (in this case, 6 cities),
• $C(n, k)$ represents the number of combinations.

Substituting the values:

$C(13, 6) = \frac{13!}{6!(13-6)!} = \frac{13!}{6!7!}$

Now let's calculate:

$C(13, 6) = \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{1235520}{720} = 1716$

Thus, the number of different selections of cities is 1716.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the calculation change if the firm could only choose 4 cities instead of 6?
2. What if the firm could select all 13 cities—how many ways would that be?
3. How do combinations differ from permutations in terms of selection?
4. If the cities were to be ranked in order, how many possible ordered selections of 6 cities would there be?
5. What real-world factors might influence the decision of how many cities to expand into?

Tip: In combination problems, order doesn't matter. But if the problem involves ranking or sequencing, you should use permutations instead.