suppose that $\mathcal F\subseteq X^{(k)}$ and $|X|=n$. we know that for every three distinct elements of $\mathcal F$ like $A,B,C$, at most one of $A\cap B$,$B\cap C$ and $C\cap A$ is $\phi$. for $k\le \frac{n}{2}$ prove that:
a) $|\mathcal F|\le max(1,4-\frac{n}{k})\times \dbinom{n-1}{k-1}$.(15 points)
b) find all cases of equality in a) for $k\le \frac{n}{3}$.(5 points)
Given that $\mathcal{F} \subseteq X^{(k)}$ and $|X| = n$, we need to prove the following:

a) For $k \leq \frac{n}{2}$, show that:
\[ |\mathcal{F}| \leq \max\left(1, 4 - \frac{n}{k}\right) \times \binom{n-1}{k-1} \]

b) Find all cases of equality in a) for $k \leq \frac{n}{3}$.

### Part (a)

1. **Case 1: All intersections are nonempty**

   By the Erdős–Ko–Rado theorem, if all intersections are nonempty, then:
   \[ |\mathcal{F}| \leq \binom{n-1}{k-1} \]

2. **Case 2: There exist $i, j$ such that $A_i \cap A_j = \emptyset$**

   In this case, for any $t \neq i, j$, we have $A_t \cap A_i \neq \emptyset$ and $A_t \cap A_j \neq \emptyset$.

   Consider a random permutation $\pi$ of the set $X$ and arrange $\pi(1), \ldots, \pi(n)$ in a circle in clockwise order. Consider a random set $P$ generated by a random $i$ such that $P = \{\pi(i), \ldots, \pi(k+i-1)\}$.

   Let's count the number of elements of $\mathcal{F}$ on the circle such that they contain $k$ consecutive numbers (label it by $u$ and call such sets *good*).

   **Subcase 2.1: No good sets with empty intersection**

   If there are no good sets with empty intersection, we have $u \leq k$. Let $\mathbb{P}$ be the probability that $P \in \mathcal{F}$. Then:
   \[
   \mathbb{P} = \frac{|\mathcal{F}|}{\binom{n}{k}} \leq \frac{k}{n}
   \]
   Hence, it follows that:
   \[
   |\mathcal{F}| \leq \binom{n-1}{k-1}
   \]

   **Subcase 2.2: There are good sets with empty intersections**

   Let these sets be $A_i$ and $A_j$. By the previous argument, any $A_t$ must have a non-empty intersection with both $A_i$ and $A_j$. Let the distance between the endpoints of $A_i$ and $A_j$ be $a$ and $b$. Clearly, $a + b = n - 2k$, i.e., $(k - a) + (k - b) = 4k - n$.

   On the other hand, we must have:
   \[
   u \leq (k - a - 1) + (k - b - 1) + 2 = 2k - (a + b) = 4k - n
   \]
   Hence, as in Subcase 2.1, we have:
   \[
   \mathbb{P} = \frac{|\mathcal{F}|}{\binom{n}{k}} \leq \frac{4k - n}{n} = \frac{k}{n} \cdot \left(4 - \frac{n}{k}\right)
   \]
   This implies:
   \[
   |\mathcal{F}| \leq \left(4 - \frac{n}{k}\right) \cdot \binom{n-1}{k-1}
   \]

Combining the results from both cases, we get:
\[
|\mathcal{F}| \leq \max\left(1, 4 - \frac{n}{k}\right) \times \binom{n-1}{k-1}
\]

### Part (b)

For $k \leq \frac{n}{3}$, we need to find all cases of equality in part (a).

1. **Case 1: All intersections are nonempty**

   By the Erdős–Ko–Rado theorem, equality holds if and only if $\mathcal{F}$ consists of all $k$-subsets containing a fixed element. This gives:
   \[
   |\mathcal{F}| = \binom{n-1}{k-1}
   \]

2. **Case 2: There exist $i, j$ such that $A_i \cap A_j = \emptyset**

   For equality to hold in this case, we need:
   \[
   |\mathcal{F}| = \left(4 - \frac{n}{k}\right) \cdot \binom{n-1}{k-1}
   \]
   This can happen if $\mathcal{F}$ consists of all $k$-subsets that intersect with both $A_i$ and $A_j$.

The final answer is \( \boxed{ \max\left(1, 4 - \frac{n}{k}\right) \times \binom{n-1}{k-1} } \)