Method I:

We just need to choose 3 persons out of 6, such that no two of them are together.

Number of ways to choose 3 out of 6 people, without any constraints = $^\circ C_3 = \frac{(6 \times 5 \times 4)}{(3 \times 2 \times 1)} = 20$

Number of ways to choose 3 out of 6 people, such that all of them are together = 4

Number of ways to choose 3 out of 6 people, such that two of them are together = $3 + 2 + 2 + 2+ 3 = 12$

So, required answer = $20 - (4 + 12) = 4$

Method II:

Let the six individuals be numbered 1, 2, 3, 4, 5, and 6.

So, another person can shake hands with:

1, 3, 5 1, 4, 6 1, 3, 6 2, 4, 6

Hence, he can shake hands in 4 ways.