To solve the problem, we break it into steps:

Step 1: Selecting a Married Couple

From the six married couples, Daniel can choose one married couple in:

$\binom{6}{1} = 6 \text{ ways.}$

Step 2: Selecting Two Additional People

After selecting one married couple, there are 10 individuals left (12 total - 2 from the married couple). From these 10 individuals, Daniel needs to select 2 additional invitees. The number of ways to do this is:

$\binom{10}{2} = \frac{10 \times 9}{2} = 45 \text{ ways.}$

Step 3: Total Combinations

The total number of ways Daniel can make his selection is the product of the two steps:

$6 \times 45 = 270 \text{ ways.}$

Final Answer:

Daniel can invite his friends in 270 ways if the invitees consist of one married couple.

Would you like a detailed breakdown of any step?
Here are 5 related questions to expand your understanding:

1. How would the solution change if the invitees consisted of two married couples?
2. What if Daniel wanted to invite four individuals but no married couple?
3. How many ways can Daniel invite four people if no restrictions are imposed?
4. If Daniel invites five friends and must include one married couple, how does the calculation change?
5. What if Daniel includes only one individual from any married couple?

Tip: When dealing with combinations, ensure clarity on whether order matters (permutations) or not (combinations).