There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not

Question

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?

Accepted Answer

Let the 6 persons be P1, P2, P3, P4, P5, P6. We need to choose 3 persons such that no two chosen persons are consecutive.

This is a classic combinatorial problem of selecting 'k' non-consecutive items from 'n' items arranged in a row. The formula for this is C(n - k + 1, k), where C denotes combinations.

In this problem:
n = 6 (total number of persons)
k = 3 (number of persons to be chosen)

Applying the formula:
Number of combinations = C(6 - 3 + 1, 3)
= C(4, 3)

Calculating C(4, 3):
C(4, 3) = 4! / (3! * (4-3)!)
= 4! / (3! * 1!)
= (4 * 3 * 2 * 1) / ((3 * 2 * 1) * 1)
= 4

The 4 distinct possible combinations are:
1.  (P1, P3, P5)
2.  (P1, P3, P6)
3.  (P1, P4, P6)
4.  (P2, P4, P6)

Thus, there are 4 distinct possible combinations.

The final answer is B

Answer

3

Answer

4

Answer

5

Answer

6

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?

Explanation

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