Note that $S\equiv \{0,1,2,\dots ,21\} \pmod4$, and in $S$ are exactly $6$ numbers $0 \pmod 4$, $6$ numbers $1 \pmod 4$, $5$ numbers $2 \pmod 4$ and $5$ numbers $3\pmod 4$. A red subset of $S$ with $4$ elements can be formed in these ways $4$ numbers of the form $4k+1$ or $4k+3$ $3$ numbers of the form $4k+1$ or $4k+3$ and the other number is of the form $4k+2$ or $4k$ $2$ numbers of the form $4k+1$ or $4k+3$ and two other numbers are of the form $4k$ and $4k+2$ Counting these we get $x=\dbinom{6}{4}+\dbinom{5}{4}+11 \left(\dbinom{6}{3}+\dbinom{5}{3} \right)+30\left(\dbinom{6}{2}+\dbinom{5}{2}\right)=1100$ A red subset of $S$ with $5$ elements can be formed in these ways $5$ numbers of the form $4k+1$ or $4k+3$ $4$ numbers of the form $4k+1$ or $4k+3$ and the other number is of the form $4k+2$ or $4k$ $3$ numbers of the form $4k+1$ or $4k+3$ and two other numbers are of the form $4k$ and $4k+2$ Counting these we get $y=\dbinom{6}{5}+\dbinom{5}{5}+11\left(\dbinom{6}{4}+\dbinom{5}{4}\right)+30\left(\dbinom{6}{3}+\dbinom{5}{3} \right)=1127$ So we have $x<y$