In how many ways can 1, 2, 3, 4, 5 and 6 be arranged in a row so that thedifference between any two adjacent numbers is not equal to 3?

This problem can be solved by using the concept of permutations and combinations.

Step 1: Identify the total number of arrangements without any restrictions.

The total number of ways to arrange 6 distinct numbers is 6!, which equals 720.

Step 2: Identify the arrangements that violate the condition.

We need to find the number of arrangements where the difference between any two adjacent numbers is equal to 3. The pairs that have a difference of 3 are (1,4), (2,5), and (3,6).

For each pair, consider them as a single entity. Now, we have 5 entities to arrange, which can be done in 5! = 120 ways. But within each entity, the two numbers can be arranged in 2! = 2 ways. So, for each pair, the total number of arrangements is 120 * 2 = 240.

Since we have 3 such pairs, the total number of arrangements that violate the condition is 240 * 3 = 720.

Step 3: Subtract the violating arrangements from the total arrangements.

The number of arrangements that satisfy the condition is the total arrangements minus the violating arrangements, which is 720 - 720 = 0.

So, there are 0 ways to arrange the numbers 1, 2, 3, 4, 5, and 6 in a row such that the difference between any two adjacent numbers is not equal to 3.