To solve this problem, we can use the concept of combinations and the principle of inclusion-exclusion.

Given:
- Number of friends = 10
- Number of guests at the dinner party = 6
- 2 friends will not attend the party together

First, let's calculate the total number of ways to select 6 guests from 10 friends without any restrictions.

Total ways to select 6 guests from 10 friends = C(10, 6)
= 10! / (6! * (10-6)!)
= 10! / (6! * 4!)
= (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1)
= 210

Now, let's consider the restriction that 2 friends will not attend the party together. We can use the principle of inclusion-exclusion to calculate the number of ways to select the guests while satisfying this restriction.

Let's assume the two friends who cannot attend the party together as A and B.

Step 1: Calculate the number of ways to select 6 guests without considering the restriction.
= Total ways to select 6 guests from 10 friends
= 210

Step 2: Calculate the number of ways to select 6 guests where both A and B are present.
We need to select 4 guests from the remaining 8 friends (excluding A and B).
= C(8, 4)
= 8! / (4! * (8-4)!)
= (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1)
= 70

Step 3: Calculate the number of ways to select 6 guests where both A and B are absent.
We need to select 6 guests from the remaining 8 friends (excluding A and B).
= C(8, 6)
= 8! / (6! * (8-6)!)
= (8 * 7) / (2 * 1)
= 28

Step 4: Calculate the number of ways to select 6 guests where either A or B is absent.
= Total ways to select 6 guests from 10 friends - Number of ways where both A and B are present - Number of ways where both A and B are absent
= 210 - 70 - 28
= 112

Therefore, the number of ways to select 6 guests from among 10 friends, where 2 friends will not attend the party together, is 112 (option C).