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? 

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.
