Express Composite Numbers to Primes

Question

Let: $p$- prime, $k$-even, $c$-composite.
$$ \begin{eqnarray}\label{eq:25d} \nonumber 3k \choose 2k &=& \frac{(2k+1)(2k+2)(2k+3)(2k+4)(2k+5)(2k+6)\ldots (2k+k)}{ k!}\\ \nonumber &=& \dfrac{1}{k!} \prod_{m=2k+1}^{3k} {m} \\ \label{eq: 25g} &=& \dfrac{1}{k!} \prod_{2k < m \leq 3k}{m}\\ \nonumber &=& \dfrac{1}{k!} \left[\prod_{2k < p \leq 3k} {p} \cdot \prod_{2k < c \leq 3k} {c} \right] \\ &=& \left[\dfrac{1}{k!} \prod_{2k < c \leq 3k} {c} \right] { \prod_{2k < p \leq 3k} {p}} \\ &=& A { \prod_{2k < p \leq 3k} {p}} . \end{eqnarray} $$

I want to prove that $A \in Z^+\ $ so that ${\displaystyle \prod_{2k < p \leq 3k} {p}} $ divides $3k \choose 2k $. I can't find any theorem about consecutive composite integer. I only know that $n!= \displaystyle \prod_{p \leq n}{p^{k(n,p)}}$ where $k(n,p) = \displaystyle \sum _{t=1} ^ r\left\lfloor \frac{n}{p^t}\right\rfloor$. It's for my thesis.

Answer 1

$$\binom{3k}{2k} =\binom{3k}{k} = \frac{3k \cdot (3k-1) \cdot (3k-2) \cdots (2k + 2) \cdot (2k+1)}{k \,\,\,\cdot\, \,\,(k-1) \,\,\,\cdot \,\,\,(k-2) \,\,\,\,\,\,\cdot \,\,\cdot\,\,\cdot\,\,\,\,\, 2 \,\,\,\cdot \,\,\, 1}$$

The primes $p$ that are between $2k$ and $3k$ are all in the numerator, and none of these primes appear in the denominator. Since binomial coefficients are integers, it follows that each prime $p$ between $2k$ and $3k$ must divide $\binom{3k}{2k}$. The product of all of these primes must also divide $\binom{3k}{2k}$.