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Find the number of ways of arranging the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 in a row so that the product of any two adjacent numbers is even or a mutliple of 3

Guest Mar 6, 2023

#1**0 **

We can approach this problem by first noting that in order for the product of two adjacent numbers to be even or a multiple of 3, at least one of the numbers must be even or a multiple of 3. Thus, we can group the numbers into two sets: even numbers and multiples of 3, and odd numbers that are not multiples of 3.

There are 5 even numbers (2, 4, 6, 8, 10) and 3 multiples of 3 (3, 6, 9), so there are 8 numbers in the first set. We can arrange these numbers in 8! ways.

There are 2 odd numbers that are multiples of 3 (3, 9) and 3 odd numbers that are not multiples of 3 (1, 5, 7), so there are 5 numbers in the second set. We can arrange these numbers in 5! ways.

Now, we need to consider the placement of the second set of numbers (odd numbers). Since the product of two adjacent numbers must be even or a multiple of 3, the odd numbers can only be placed in the spaces between the even numbers and multiples of 3.

There are 9 spaces between the 8 even numbers and multiples of 3. We need to choose 5 of these spaces to place the 5 odd numbers. We can do this in 9 choose 5 ways (or 9!/5!4!).

**Therefore, the total number of arrangements is 8! × 5! × [9 C 5] =609,638,400 .**

Guest Mar 6, 2023