Another question about ratios of Pochhammer symbols My question is similar to this question. Can
$$\frac{(11/6)_n (7/6)_n (3/2)_n}{(3)_n}$$
be expressed 'nicely' in terms of factorials just like $(1/6)_n (1/2)_n (5/6)_n$ in the aforementioned question?
Is there a general formula which allows one to convert from Pochhammer symbols to factorials?
 A: Using the key formula for rising factorials of rational numbers 

$$\left(\frac{a}b\right)_n=\frac1{b^n}\prod_{k=1}^n(bk+a-b),$$ 

one sees that the numerator of the ratio $R_n$ to be computed is $$\left(\frac{11}6\right)_n\left(\frac{9}6\right)_n\left(\frac{7}6\right)_n=\frac1{6^{3n}}\prod_{k=1}^n(6k+5)(6k+3)(6k+1)=\frac1{6^{3n}}\,(\ast).$$ The product $(\ast)$ on the RHS enumerates every odd integer from $7$ to $6n+5$ hence, inserting every even number from $8$ to $6n+6$ included, one sees that $$(\ast)=\frac{(6n+6)!}{6!}\prod_{k=4}^{3n+3}\frac1{2k}=\frac{(6n+6)!}{6!}\frac{3!}{2^{3n}(3n+3)!}.$$
On the other hand, $$(3)_n=\frac{(n+2)!}2,$$ hence the whole ratio to be computed is
$$
R_n=\frac1{6^{3n}}\frac{(6n+6)!}{6!}\frac{3!}{2^{3n}(3n+3)!}\frac2{(n+2)!}=\frac1{12^{3n}}\frac{(6n+6)!}{(3n+3)!(n+2)!}\frac{3!2!}{6!}.$$ Using the identity
$$
\frac{(a+b)!}{b!}=(b+1)_a,
$$
for various integers $(a,b)$, one gets finally the pair of equivalent formulas

$$R_n=\frac1{60}\frac1{12^{3n}}\frac{(6n+6)!}{(3n+3)!(n+2)!}=\frac1{12^{3n}}\frac{(7)_{6n}}{(4)_{3n}(3)_n}.$$

A: First of all it is the first answer to give what the OP asked for. 
What I gave you is correct. It is a convention issue. However see what I added. The main thing in your question is to express the Pochhammer symbol in terms of factorials and it was given in this answer. The rest is a convention issue and you can move from one to another as you can see in the add.
Here is what you need

$$ (x)_n=\frac{\Gamma(x+1)}{\Gamma(x-n+1)}, $$

where $\Gamma(x+1)=x!$ is the generalization of the factorial.  For instance

$$ (11/6)_n = \frac{\Gamma(17/6)}{\Gamma(17/6-n)}. $$

See (I), (II).
Added: You can use the relation between the falling and rising factorials

$$ x^{(n)} = {(-1)}^n {(-x)}_{{n}}  $$

where $ x^{(n)}  $ is the rising factorial

$$ x^{(n)} = \frac{\Gamma(x+n)}{\Gamma(x)}.  $$  

A: Note that:
$$
\left(\frac{11}{6}\right)_n = \left(\frac{11}{6}\right)\left(\frac{5}{6}\right)_{n-1} \\
\left(\frac{7}{6}\right)_n = \left(\frac{7}{6}\right)\left(\frac{1}{6}\right)_{n-1} \\
\left(\frac{3}{2}\right)_n = \left(\frac{3}{2}\right)\left(\frac{1}{2}\right)_{n-1} 
$$
With this, it should be straightforward.
