# 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?

2 years, 3 months ago

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- 1 year, 12 months ago

- 1 year, 12 months ago

- 1 year, 12 months ago

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.

- 2 years, 3 months ago

Ah...

I forgot to divide by 2! in my first case to remove the ordering. Thanks for the clarification.

- 2 years, 3 months ago

Can a number be repeated in the pairs?

- 2 years, 3 months ago

No.

- 2 years, 3 months ago

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

Still working on Case 2.

- 2 years, 3 months ago

Case 2:$$4!$$.

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

- 2 years, 3 months ago

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?

- 2 years, 3 months ago

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.

- 2 years, 3 months ago

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?

- 2 years, 3 months ago

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.

- 2 years, 3 months ago

Yes,correct.It should be 864 had the order of selection mattered.

- 2 years, 3 months ago

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?

- 2 years, 3 months ago

You have added the number of ways of case 1 and case 2 incorrectly.Even by your method the answer is 36.

- 2 years, 3 months ago

Comment deleted Apr 30, 2016

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)}.

- 2 years, 3 months ago

Sorry.

- 2 years, 3 months ago

No problem. I have editted the problem so as to ensure clarity of the problem statement.

- 2 years, 3 months ago

Now it's better

- 2 years, 3 months ago