prove that the expression $\frac{(3n)!}{(3!)^n}$ is integral for $n \geq 0$

Question

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?

Answer 1

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 $n!$.

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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 $n!$

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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.

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Solution $3$:

$$\frac{(3n)!}{n!^3}=\binom{3n}{n,n,n}$$

Look up multinomial coefficient.

Answer 2

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.

Answer 3

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 $(3!)^n = 3^n \cdot 2^n$.

Answer 4

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 $6^n \mid (3n)!$

Answer 5

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 !

Answer 6

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.