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Problem: Let $S=\{1,2,...,8\}$ be the set consisting first eight positive integer from which each integer present can be taken twice not more than that. Select a quadruple $(a,b,c,d)$ from $S$, where $a,b,c,d$ all are distinct and the sum of $a+b+c+d = \text{odd integer}$. Is it possible to find four such quadruples?

For Example: Consider $S=\{1,2,3,4,5,6\}$ and the pairs $(1,2,4,5)[sum=12],(2,3,5,6)[sum=16]$ and $(1,3,4,6)[sum=14]$ in this example the sum of each integer and each triplet is even integer. In above problem, I need such four quadruple for which the sum must be odd integer.

Kindly give some Hint or solution to this problem. Thanks in advance.

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    $\begingroup$ Four integers hardly form a pair ;) $\endgroup$
    – Hagen von Eitzen
    Commented Mar 31, 2020 at 10:04
  • $\begingroup$ @HagenvonEitzen Do you mean pair is not the suitable word for presenting four integer $(a,b,c,d)$? $\endgroup$
    – Siddhant Trivedi
    Commented Mar 31, 2020 at 10:09
  • $\begingroup$ "Pair" means two (ordered) objects; e.g., a pair of shoes consists of 1) the left shoe and 2) the right shoe. For larger ordered collection, we use names (that are not part of everyday English) such as triple for three objects, quadruple for four objects, etc. So I guess you want quadruple ... $\endgroup$
    – Hagen von Eitzen
    Commented Mar 31, 2020 at 10:11
  • $\begingroup$ @HagenvonEitzen, I believe what the OP wants is four $4$-tuples, $(a_i,b_i,c_i,d_i)$ with $1\le a_1\lt b_i\lt c_i\lt d_i\le8$ for $1\le i\le 4$ such that each number $1\le k\le 8$ appears twice (and $a_i+b_i+c_i+d_i$ is odd for all four $i$). Note, the sum of the sums must be $72$, since $1+2+\cdots+8=36$. $\endgroup$
    – Barry Cipra
    Commented Mar 31, 2020 at 10:19
  • $\begingroup$ @HagenvonEitzen Thanks for the suggestion. I have made correction according to your comment please check. Is it look good? $\endgroup$
    – Siddhant Trivedi
    Commented Mar 31, 2020 at 10:19

2 Answers 2

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I think you mean four quadruples (since they are tuples of four integers). Moreover, we are allowed to choose each integer from $1$ to $8$ atmost twice. This gives us a maximum of $16$ numbers. However, four quadruples is $4 \cdot 4 = 16$ integers. Thus, every number from $1$ to $8$ is to be used twice. It is easy to do this.

One way to construct this is that since $1+2+ \cdot +8=36$ is even, a quadruple's complement also satisfies the requirements. We only need to find two quadruples, say $\{1,2,3,5\},\{1,2,3,7\}$ and use their complements $\{4,6,7,8\},\{4,5,6,8\}$. Thus, these four quadruples satisfy the requirements.

OP requested for quadruples with sum $15,17,19,21$ in the comments. Note that: $$1+2+3+ \cdot +8 = 36 = 15+21 = 17+19$$

Thus, we need a quadruple with sum $15$ (complement will have sum $21$). We also need a quadruple with sum $17$ (complement will have sum $19$). We can choose the quadruple $\{1,2,4,8\}$ for sum $15$ whose complement will be $\{3,5,6,7\}$ with sum $21$. We can choose quadruple $\{1,2,6,8\}$ for sum $17$ whose complement will be $\{3,4,5,7\}$ with sum $19$.

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  • $\begingroup$ Can you give me quadruples whose sum is $15,17,19$ and $21$? $\endgroup$
    – Siddhant Trivedi
    Commented Mar 31, 2020 at 10:26
  • $\begingroup$ Great Answer! Thank You. $\endgroup$
    – Siddhant Trivedi
    Commented Mar 31, 2020 at 10:47
  • $\begingroup$ The way you explained would help me a lot. $\endgroup$
    – Siddhant Trivedi
    Commented Mar 31, 2020 at 10:51
  • $\begingroup$ It is my pleasure :) $\endgroup$
    – Haran
    Commented Mar 31, 2020 at 10:51
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There are $6\choose 2$ ways to pick three distinct numbers from $\{3,4,5,6,7,8\}$. By adding either $1$ or $2$, we will get four distinct integers with odd sum. This method alone already gives is $15$ solutions. There are a few more where both $1$ and $2$ or neither of them is in our selection, but we have met the goal already.

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