There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place?

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 = 6C3 = (6 × 5 × 4)/(3 × 2 × 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.