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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

 Apr 14, 2023
 #1
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To 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 multiple of 3, we can use the principle of inclusion-exclusion. First, we can count the number of ways of arranging the numbers without any restrictions. There are 10! ways to do this. Next, we can count the number of ways of arranging the numbers where the product of some pair of adjacent numbers is neither even nor a multiple of 3. There are three cases to consider: Case 1: The pair consists of two odd numbers. There are 5 pairs of odd numbers that are adjacent in the list: (1,3), (3,5), (5,7), (7,9), and (9,10). For each pair, there are 8! ways to arrange the remaining 8 numbers (since we can treat the pair as a single number and arrange the remaining numbers around it). Therefore, there are 5 x 8! ways to arrange the numbers with a pair of adjacent odd numbers. Case 2: The pair consists of an even number and a number that is not a multiple of 3. There are 5 even numbers and 4 numbers that are not multiples of 3 that can be paired with an even number. For each pair, there are 8! ways to arrange the remaining 8 numbers. Therefore, there are 5 x 4 x 8! ways to arrange the numbers with a pair of an even number and a number that is not a multiple of 3. Case 3: The pair consists of two numbers that are multiples of 3. There are 3 pairs of multiples of 3 that are adjacent in the list: (3,6), (6,9), and (9,10). For each pair, there are 8! ways to arrange the remaining 8 numbers. Therefore, there are 3 x 8! ways to arrange the numbers with a pair of adjacent multiples of 3. However, we have overcounted the arrangements where there are two or more pairs of adjacent numbers that violate the conditions. To correct for this, we need to subtract the number of arrangements with two pairs of adjacent numbers that violate the conditions, and then add back the number of arrangements with three pairs of adjacent numbers that violate the conditions. There are no arrangements with four or more pairs of adjacent numbers that violate the conditions, since this would require at least 8 numbers that are not multiples of 3, which is impossible. For two pairs of adjacent numbers that violate the conditions, there are two cases to consider: Case 1: Both pairs consist of two odd numbers. There are 4 ways to choose the pairs: (1,3) and (5,7), (1,3) and (7,9), (3,5) and (7,9), and (5,7) and (9,10). For each pair of pairs, there are 7! ways to arrange the remaining 7 numbers. Therefore, there are 4 x 7! ways to arrange the numbers with two pairs of adjacent odd numbers. Case 2: One pair consists of two odd numbers, and the other pair consists of an even number and a number that is not a multiple of 3. There are 10 ways to choose the even number, and 3 ways to choose the number that is not a multiple of 3. There are 5 ways to choose which of the two pairs contains the even number. For each choice of even number and pair, there are
 Apr 14, 2023
 #2
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10!  -  [9! * 6] + [8! * 6] ==1,693,440 - total number of ways

 Apr 14, 2023
 #3
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I'm pretty sure it's nine

 Apr 14, 2023

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