Reasoning about combinations of distinct modular classes of $5$ and $7$

Question

Let $S = \{ x \in \ Z\, | \, 1 \le x \le 35\}$

For integers $a,b,c,d$ with $0 \le a < b < 5$ and $0 \le c < d < 7$, let:

$$C_{a,b,c,d} = \{ x \in S\,|\, x \not\equiv a \pmod 5,\, x \not\equiv b \pmod 5,\, x \not\equiv c \pmod 7,\, x \not\equiv d \pmod 7 \,\}$$

There are ${ 7 \choose 2 } \times { 5 \choose 2 } = 210$ such distinct sets $C_{a,b,c,d}$

For any $1 \le x \le 34$, there are:

• ${ 6 \choose 2} \times {4 \choose 2} =90$ distinct sets $C_{a,b,c,d}$ where $x \in C_{a,b,c,d}$
• $210 - 90 = 120$ distinct sets $C_{a,b,c,d}$ where $x \not\in C_{a,b,c,d}$
• ${ 5 \choose 2} \times { 3 \choose 2} = 30$ distinct sets $C_{a,b,c,d}$ where $x \in C_{a,b,c,d}$ and $(x+1) \in C_{a,b,c,d}$

Here's my question. I am unclear about how to handle this case:

How many distinct sets $C_{a,b,c,d}$ such that $x \not\in C_{a,b,c,d}$ and $(x+1) \not\in C_{a,b,c,d}$?

I am hoping that someone can help me to understand how to count this number of distinct sets.

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Edit:

I have restated my question based on Quasi's feedback.

Answer 1

It will be more convenient to work mod $35$. The counts will be the same.

Applying that change, let

• $Z_{35}=\{0,...,34\}$.$\\[4pt]$
• $Z_{5}=\{0,...,4\}$.$\\[4pt]$
• $Z_{7}=\{0,...,6\}$.

For $a,b\in Z_5$ with $0\le a < b$, and $c,d\in Z_7$ with $0\le c < d$, let $$ C_{a,b,c,d} = \{ x \in Z_{35}\mid x \not\equiv a,b\;(\text{mod}\;5), x \not\equiv c,d\;(\text{mod}\;7) \} $$ Claim:$\;$For all $x\in Z_{35}$, there are exactly $60$ sets $C_{a,b,c,d}$ such that $x,x+1\not\in C_{a,b,c,d}$.

Consider the complementary count . . .

Let $x\in Z_{35}$.

• There are exactly $90$ sets $C_{a,b,c,d}$ such that $x\in C_{a,b,c,d}$.$\\[4pt]$
• There are exactly $90$ sets $C_{a,b,c,d}$ such that $x+1\in C_{a,b,c,d}$.$\\[4pt]$
• There are exactly $30$ sets $C_{a,b,c,d}$ such that $x,x+1\in C_{a,b,c,d}$.

hence there are $90+90-30=150$ sets $C_{a,b,c,d}$ such that at least one of $x,x+1$ is in $C_{a,b,c,d}$.

Since there are a total of $210$ distinct sets $C_{a,b,c,d}$, it follows that there are exactly $210-150=60$ sets $C_{a,b,c,d}$ such that $x,x+1\not\in C_{a,b,c,d}$.