How many winning hands are possible in Mahjong? A set of Mahjong tiles consists of four copies of 34 unique tiles (disregard "Bonus" tiles for the purpose of this question):

There are three ways of creating a meld:

*

*Eye: two copies of the same tile

*Pong: three copies of the same tile

*Kong: four copies of the same tile

*Chow: three consecutive tiles (only possible for each of the suits)

A winning hand consists of exactly one eye and four non-eye melds. A tile cannot be shared across multiple melds.
My attempt at calculating an upper limit for the number of different winning hands is below:
34 unique tiles gives 34 unique eyes, 34 unique pongs and 34 unique kongs. Within a single suit, 7 $[(1,2,3), (2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9)]$ chows are possible, giving 21 chows in total.
Choosing one pair and four kongs gives $34\times nCr(34+34+21, 4)= 83015284$
However, I realised an error in my calculations is that I assumed that a meld cannot be used twice, which is false as any chow can occur a maximum of four times. Therefore, the number above may not be the upper limit.
Could someone help with calculating how many winning hands are possible in Mahjong?
 A: An upperbound
The winning hand must contain $14$ tiles and so we have the following possibilities. I've explained the reasoning in the most difficult case of the four Chows.
Eye and three Kongs $$\begin{pmatrix}34\\4\\\end{pmatrix} \times 4=185504.$$
Eye and four Pongs $$\begin{pmatrix}34\\5\\\end{pmatrix} \times 5=1391280.$$
Eye, Chow and three Pongs $$21\times\begin{pmatrix}34\\4\\\end{pmatrix} \times 4=3895584.$$
Eye, two Chows and two Pongs $$\left (\begin{pmatrix}21\\2\\\end{pmatrix}+21\right)\times\begin{pmatrix}34\\3\\\end{pmatrix} \times 3=4146912.$$
Eye, three Chows and one Pong $$\left (\begin{pmatrix}21\\3\\\end{pmatrix}+\begin{pmatrix}21\\2\\\end{pmatrix}\times 2+21\right) \times\begin{pmatrix}34\\2\\\end{pmatrix} \times 2=1751442.$$
Eye and four Chows
The Chows can be all different $\begin{pmatrix}21\\4\\\end{pmatrix}$.
There can be three different Chows $\begin{pmatrix}21\\3\\\end{pmatrix}\times3$.
There can be two different Chows $\begin{pmatrix}21\\2\\\end{pmatrix}\times3$.
The Chows can be all the same $\begin{pmatrix}21\\1\\\end{pmatrix}$.
This adds up to $10626$ possibilities.
There is then a maximum of $34$ possibilities for the Eye.
$$10626\times34=361284.$$
Total
$$11,732,006$$
N.B.
I hope I've understood the rules correctly!
This answer is an upper bound. There will be cases where choosing overlapping Chows will limit the possibilities for the other tiles but these will be relatively few in number. For example, in the case of four identical Chows there will only be $31$ possibilities for the Eye. Thus our count has been $3\times 21=63$ too large for these possibilities.
