There are 6 persons arranged in a row. Another person has to snake 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?

The correct option is 4.

Step-by-step Explanation:
We are given a row of 6 persons. Let us label these 6 persons in order as P₁, P₂, P₃, P₄, P₅, and P₆.
Another person needs to shake hands with exactly 3 of these persons such that no two chosen persons are standing consecutively in the row.

Method 1: Formula for Selecting Non-Consecutive Objects
The total number of ways to choose $k$ non-adjacent items from $n$ items arranged in a straight line is given by the formula:

$\text{Number of combinations}=C(n-k+1,k)$

Here, the total number of persons is:

$n=6$

The number of persons to select for handshakes is:

$k=3$

Substituting these values into the expression $n-k+1$:

$6-3+1=4$

Now, calculate the combination $C(4,3)$:

$C(4,3)=\frac{4!}{3!\times (4-3)!}$

$C(4,3)=\frac{4\times 3\times 2\times 1}{(3\times 2\times 1)\times 1}=4$

Method 2: Direct Listing of Valid Combinations
We can also list all possible sets of 3 non-consecutive persons chosen from {P₁, P₂, P₃, P₄, P₅, P₆}:
1. (P₁, P₃, P₅)
2. (P₁, P₃, P₆)
3. (P₁, P₄, P₆)
4. (P₂, P₄, P₆)

There are no other valid selections of 3 persons where no two are adjacent. Therefore, exactly 4 distinct handshake combinations can take place.