To solve this problem, we need to calculate the number of ways Mawia can invite 6 friends out of 10 such that 3 particular friends are never included.

Step 1: Total number of ways to invite 6 friends from 10

The total number of ways to select 6 friends out of 10 is given by the combination formula:

$\text{Total ways} = \binom{10}{6} = \frac{10!}{6! \cdot (10-6)!} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Step 2: Number of ways to invite 6 friends such that 3 particular friends are not included

If 3 particular friends are never attending, Mawia has only 7 friends left to choose from. So, we need to calculate the number of ways to select 6 friends from these 7:

$\text{Ways without the 3 particular friends} = \binom{7}{6} = \frac{7!}{6! \cdot (7-6)!} = \binom{7}{1} = 7$

So, there are 7 ways in which the 3 particular friends are never invited to the party.

Final Answer:

There are 7 ways in which the 3 particular friends never attend the party.

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Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How many ways can Mawia invite exactly 3 of the 3 particular friends?
2. How many ways can Mawia invite at least one of the 3 particular friends?
3. How many ways can Mawia invite all 3 of the particular friends?
4. If Mawia decided to invite 7 friends instead, how many ways could 3 particular friends never be invited?
5. How many ways can Mawia invite only 2 of the 3 particular friends?

Tip: When dealing with combinatorial problems, always identify if certain elements must be included or excluded to simplify the problem.