Find the number of ways of arranging the numbers 1,2 ,3, 4, 5, 6 in a row so that the product of any two adjacent numbers is even.
There are 6! = 720 ways to arrange the numbers 1, 2, 3, 4, 5, 6 in a row. However, not all of these arrangements will have the product of any two adjacent numbers being even.
In order for the product of two adjacent numbers to be even, at least one of the numbers must be even. Therefore, we can group the numbers into two sets: even numbers and odd numbers. There are 3 even numbers (2, 4, 6) and 3 odd numbers (1, 3, 5).
We can arrange the even numbers in 3! = 6 ways. We can arrange the odd numbers in 3! = 6 ways. Now, we need to consider the placement of the odd numbers. Since the product of two adjacent numbers must be even, the odd numbers can only be placed in the spaces between the even numbers. There are 4 spaces between the 3 even numbers. We need to choose 3 of these spaces to place the 3 odd numbers. We can do this in 4C3 = 4 ways.
Therefore, the total number of ways of arranging the numbers 1, 2, 3, 4, 5, 6 in a row so that the product of any two adjacent numbers is even is 3! * 3! * 4 = 72.
There are 3 numbers to be excluded from being next to each other:
That is: 1,3, 5 ==3! =6 permutations
There are: 6 P 2 ==6 x 5 ==30 permutations
6 / 30 x 6! ==144 ways of arranging them so that the product of any 2 is even.