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Doubt in a combinatorics problem

Suppose we have a set \(S=\{2,3,4,5,6,7,8,9\}\). In how many ways, can we select 4 pairs of numbers from \(S\) such that the greatest common divisor of the two numbers in each pair is not equal to 2?

Note by Indraneel Mukhopadhyaya
7 months, 2 weeks ago

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@Indraneel Mukhopadhyaya \ Virat Kohli · 3 months, 3 weeks ago

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(@Indraneel Mukhopadhyaya Virat Kohli · 3 months, 3 weeks ago

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@Indraneel Mukhopadhyaya Virat Kohli · 3 months, 3 weeks ago

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I was getting the answer as 36.

My cases were similar to that of Deeparaj.

Case 1: When (4,8) is one of the selected pair.

Among the remaining 6 numbers only (2,6) have GCD=2. We can select any 3 pairs from the remaining 6 numbers in ((6C2)(4C2)(2C2)/3!)=15 ways( Note that we have to only select the pairs, hence the factor of 3! in the denominator). From this we need to subtract the ways where (2,6) is one of the pairs. Hence the answer of case 1 is 15-3=12.

Case 2: When (4,8) is not of the pairs.

In this case we can show that in each of the 4 pairs we must have one odd number and one even. Therefore total number ways of selecting 4 pairs in this case is simply 4!=24. Nitish Joshi · 7 months, 1 week ago

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@Nitish Joshi Ah...

I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification. Deeparaj Bhat · 7 months, 1 week ago

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Can a number be repeated in the pairs? Deeparaj Bhat · 7 months, 1 week ago

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@Deeparaj Bhat No. Indraneel Mukhopadhyaya · 7 months, 1 week ago

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@Indraneel Mukhopadhyaya Case 1: One of the pairs is (4,8): \(4\times {4\choose2}\)

Still working on Case 2. Deeparaj Bhat · 7 months, 1 week ago

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@Deeparaj Bhat Case 2:\(4!\).

So, on the whole \(\boxed{ 48 }\) ways. Am I right?

@Indraneel Mukhopadhyaya Deeparaj Bhat · 7 months, 1 week ago

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@Deeparaj Bhat You see,I do not know the correct answer.But,according to me the answer cannot be so less.Just to prove my point,let us assume that 2 is paired with 3.So,we have our first pair (2,3) whose gcd is not 2.Now,we can make the remaining three pairs using 4,5,6,7,8,9 ; but we must ensure that the pair (4,6) or (6,8) is not selected.So,the total number of ways of selecting the remaining three pairs is (6C2)(4C2)(2C2) - 2 (4C2)(2C2)=78.But,this is only one possible case (many other cases still remain to be considered).So, the answer must be greater than 78.Does this make sense? Indraneel Mukhopadhyaya · 7 months, 1 week ago

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@Indraneel Mukhopadhyaya I think in your argument you didn't take into consideration that we have to just select the pairs. Because all the 3 remaining pairs are distinct, it would not be 78, but would be 13 instead. Nitish Joshi · 7 months, 1 week ago

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@Nitish Joshi No,it should be ((6C2)(4C2)(2C2)/3!) -2 ((4C2)(2C2))/2!=9.Also,only three other cases remain in which 2 can be paired with 5,7,9 apart from 3 whose case has been considered.The number of ways of selection for the last three cases is same as that of the first case, by symmetry.Hence,the total number of ways is 9 times (4) = 36.Is it now making sense? Would the answer be 864 had order of selection mattered? Indraneel Mukhopadhyaya · 7 months, 1 week ago

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@Indraneel Mukhopadhyaya Yes 36 seems to be correct. I too was getting the same answer but my cases were a bit different than yours. I have posted the solution. If the order of selection had mattered then answer would be 36* 4!=864. Nitish Joshi · 7 months, 1 week ago

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@Nitish Joshi Yes,correct.It should be 864 had the order of selection mattered. Indraneel Mukhopadhyaya · 7 months, 1 week ago

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@Indraneel Mukhopadhyaya I'm not very sure about your arguement.

Here's mine.

Case 1: Group 2 with any number other than 6 (since 4 and 8 are alredy grouped) and group the remaining among themselves in any manner.

Case 2: None of the evens are paired. So, the pairing of the odds with the evens can be done in 4! ways.

Edit: I get your arguement. But then, what's the flaw in mine? Deeparaj Bhat · 7 months, 1 week ago

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@Deeparaj Bhat You have added the number of ways of case 1 and case 2 incorrectly.Even by your method the answer is 36. Indraneel Mukhopadhyaya · 7 months, 1 week ago

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Comment deleted 7 months ago

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@Vignesh S I think you have interpreted the problem statement incorrectly.You do not have to select 4 numbers.You have to select 4 PAIRS of numbers such that the gcd of two numbers in each PAIR is not 2.For example, one possible selection could be {(2,9) , (3,8) , (4,7) , and (5,6)}. Indraneel Mukhopadhyaya · 7 months, 2 weeks ago

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@Indraneel Mukhopadhyaya Sorry. Vignesh S · 7 months, 2 weeks ago

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@Vignesh S No problem. I have editted the problem so as to ensure clarity of the problem statement. Indraneel Mukhopadhyaya · 7 months, 2 weeks ago

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@Indraneel Mukhopadhyaya Now it's better Vignesh S · 7 months, 2 weeks ago

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