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.
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 $19$.