prove that the expression $\frac{(3n)!}{(3!)^n}$ is integral for $n \geq 0$ My concept of real no. Is not very clear. Please also tell the logic behind the question.
The expression is true for 19, is it true for all the multiples? 
 A: You want to prove $\frac{(3n)!}{6^n}$ is an integer. Just use $\frac{3n!}{6^n}=\frac{1\cdot2\cdot3}{6}\frac{4\cdot 5\cdot 6}{6}\frac{7\cdot8\cdot 9}{6}\dots \frac{(3n-2)(3n-1)(3n)}{6}$ and each fraction is an integer since $k(k+1)(k+2)$ is always a multiple of $2$ and of $3$ since three consecutive integers always contain a multiple of $3$ and an odd number. Alternatively because the product of $n$ consecutive numbers is always divisible by $n!$.

I had misunderstood your question as prove $\frac{(3n)!}{n!^3}$ is an integer. Here is a solution to that problem.
Solution 1:
$$\frac{(3n)!}{n!^3}=\frac{1\times2\times\dots \times n}{n!}\frac{(n+1)\times (n+2)\times (n+n)!}{n!}\frac{(2n+1)\times (2n+2)\times \dots (2n+n)}{n!}$$
all three factors are integers since the product of $n$ consecutive integers is divisible by $n!$

Solution $2$:
$$\frac{(3n)!}{n!^3}=\frac{1\times2\times\dots \times n}{n!}\frac{(n+1)\times (n+2)\times (n+n)!}{n!}\frac{(2n+1)\times (2n+2)\times \dots (2n+n)}{n!}=\binom{3n}{n}\binom{2n}{n}\binom{n}{n}$$
Look up binomial coefficient.

Solution $3$:
$$\frac{(3n)!}{n!^3}=\binom{3n}{n,n,n}$$
Look up multinomial coefficient.
A: Very simply the expression is the number of ways of arranging $ 3n $ objects, of which there are $n$ distinct group of $3$ alike objects. eg.  Number of ways of arranging 12 balls of which 3 are blue, 3 are red, 3 are green and 3 are yellow.  Since this expression is the number of ways, so it is an integer alright.  More of an intuitive way of looking at it.
A: In $(3n)! = 1 \times 2 \times \ldots \times 3n$ there are $n$ numbers divisible by 3 directly ($3, 6, \ldots, 3n$) and additionally $n$ which are divisible by $2$, so $(3n)!$ is divisible by $(3!)^n = 3^n \cdot 2^n$.
A: Observe that : there are $n$ numbers which are both divisible by $2$ and $3$ that appear in the product of $(3n)! = 1\cdot 2\cdot 3\cdots (3n)$ are: $2,4,6,...,2n$, and $3,6,9,...,3n$. Thus the product is divisible by both $3^n$ and $2^n$, and since $(2^n,3^n) = 1$, we have $6^n \mid (3n)!$
A: Here is a proof by induction 
Base case $n=0$ , we have $$\frac{(3 \times 0)!}{(3!)^0} = \frac{0!}{6^0} = \frac{1}{1} = 1$$ because $0! = 1$ and $6^0 =1$ and so $1$ is an integral number so the base case works
Now assume it works for an integer $k \geq 0$ then we need to prove it  for $k+1$ , here is how
Since it work for $k$ then we have $$\frac{(3k)!}{(3!)^k} = m$$ for some integral number $m$, now consider $$\frac{(3(k+1))!}{(3!)^{k+1}}$$ which is equal to $$\color{blue}{\frac{(3k + 3)!}{3!^k \times 3!}}$$
Now $(3k+3)! = (3k +3)(3k +2)(3k+1)(3k)!$ and so now we have the blue fraction equal to $$\frac{ (3k +3)(3k +2)(3k+1)(3k)!}{3!^k \times 3!}$$ which is equal to $$\frac{ (3k +3)(3k +2)(3k+1)}{3!} \times \frac{(3k)!}{(3!)^k}$$ and remember that $$\frac{(3k)!}{(3!)^k} = m$$ and so we  have $$\frac{ (3k +3)(3k +2)(3k+1)}{3!} \times m$$ now $$\frac{ (3k +3)(3k +2)(3k+1)}{3!=6}$$ is divisible by $6$ ($\color{red}{I \space will \space  leave \space  it \space  you \space  to  \space prove  \space it \space  ! )}$ and so we have  $$\frac{ (3k +3)(3k +2)(3k+1)}{3!} \times \frac{(3k)!}{(3!)^k} = 6am$$ for some integer $a$ which is an integral number as well , hence done !
A: prove $(3n)!\over{6^n}$ is integer.  no need for induction, etc.  $6^n=2^n3^n$
$(3n)!=(3n)(3n-1)(3n-2)...(3)(2)(1)$ with $3n$ total factors.  half of those factors must be even, which takes care of $2^n$.  for $3^n$ every $3rd$ factor going back from $3n$ must be divisible by $3$:
$\{3n,3n-3,3n-6,...,9,6,3\}$
divide all of those by $3$: $\{n,n-1,n-2,...,3,2,1\}$;  there are $n$ of them.  this takes care of $3^n$
so $(2^n)(3^n)=6^n|(3n)!$, making $(3n)!\over{6^n}$ an integer. 
