$ R_k(m,n)$ is the least number such that for each coloring of $ k$-subsets of $ \{1,2,\dots,R_k(m,n)\}$ with blue and red colors, there is a subset with $ m$ elements such that all of its k-subsets are red or there is a subset with $ n$ elements such that all of its $ k$-subsets are blue.
a) If we give a direction randomly to all edges of a graph $ K_n$ then what is the probability that the resultant graph does not have directed triangles?
b) Prove that there exists a $ c$ such that $ R_3(4,n)\geq2^{cn}$.
### Part (a)
To find the probability that a randomly directed complete graph \( K_n \) does not contain any directed triangles, we need to consider the following:

1. **Total Number of Directed Edges**:
   In a complete graph \( K_n \), there are \( \binom{n}{2} \) edges. Each edge can be directed in 2 ways (either from vertex \( u \) to vertex \( v \) or from vertex \( v \) to vertex \( u \)).

2. **Total Number of Possible Directed Graphs**:
   Since each of the \( \binom{n}{2} \) edges can be directed in 2 ways, the total number of possible directed graphs is \( 2^{\binom{n}{2}} \).

3. **Number of Directed Triangles**:
   A directed triangle is a set of three vertices \( \{u, v, w\} \) such that the edges between them form a cycle (e.g., \( u \to v \to w \to u \)). For a given set of three vertices, there are 8 possible ways to direct the edges, but only 2 of these form a directed triangle.

4. **Probability Calculation**:
   The probability that a given set of three vertices does not form a directed triangle is \( 1 - \frac{2}{8} = \frac{3}{4} \).

   There are \( \binom{n}{3} \) sets of three vertices in \( K_n \). The probability that none of these sets forms a directed triangle is:
   \[
   \left( \frac{3}{4} \right)^{\binom{n}{3}}
   \]

Thus, the probability that a randomly directed complete graph \( K_n \) does not contain any directed triangles is:
\[
\boxed{\left( \frac{3}{4} \right)^{\binom{n}{3}}}
\]

### Part (b)
To prove that there exists a constant \( c \) such that \( R_3(4, n) \geq 2^{cn} \), we use a probabilistic method.

1. **Transitive Tournament**:
   A transitive tournament is a directed graph where the vertices can be ordered such that for any two vertices \( u \) and \( v \) with \( u < v \), there is a directed edge from \( u \) to \( v \).

2. **Expected Number of Ladders**:
   Consider a tournament on \( k \) vertices. The probability that a given subset of \( n \) vertices forms a transitive tournament (ladder) is \( 2^{-\binom{n}{2}} \cdot n! \).

3. **Expected Number of Size-\( n \) Ladders**:
   The expected number of size-\( n \) ladders in a tournament on \( k \) vertices is:
   \[
   \binom{k}{n} \cdot 2^{-\binom{n}{2}} \cdot n!
   \]

4. **Bounding the Expected Number**:
   We want this expected number to be less than 1 to ensure that there exists a tournament with no size-\( n \) ladders. This gives us:
   \[
   \binom{k}{n} \cdot 2^{-\binom{n}{2}} \cdot n! < 1
   \]

5. **Simplifying the Inequality**:
   Using the approximation \( \binom{k}{n} \approx \frac{k^n}{n!} \), we get:
   \[
   \frac{k^n}{n!} \cdot 2^{-\binom{n}{2}} \cdot n! < 1
   \]
   \[
   k^n \cdot 2^{-\binom{n}{2}} < 1
   \]
   \[
   k^n < 2^{\binom{n}{2}}
   \]
   \[
   k < 2^{\frac{\binom{n}{2}}{n}}
   \]
   \[
   k < 2^{\frac{n-1}{2}}
   \]

6. **Conclusion**:
   By choosing \( k = 2^{\frac{n-1}{2}} \), we ensure that there exists a tournament with no size-\( n \) ladders. Therefore, we have:
   \[
   R_3(4, n) \geq 2^{\frac{n-1}{2}}
   \]

Thus, there exists a constant \( c \) such that \( R_3(4, n) \geq 2^{cn} \).

\(\blacksquare\)