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:
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)][(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×nCr(34+34+21,4)=8301528434×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?