<s>[INST] 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? 
 [/INST] 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.
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