How many 90 ball bingo cards are there? In the UK there are 90 bingo balls. A bingo card consists of 9 columns and 3 rows. A row contains exactly five numbers and four blanks. A column consists of one, two or three numbers and never three blanks. The columns are occupied by the number chosen from 1-9 (only nine), 10-19 (ten), 20-29 ... 70-79 and importantly 80-90 (eleven). Numbers in columns are ordered from top to bottom.
How many possible unnumbered card layouts are there?
How many possible numbered cards are there?
I have started to approach this problem by first choosing the blanks for the first two rows and then attempting to work out a compensating row. This was hard.
Next I begun to understand that there were a number of equivalence classes as columns could be reordered for the purposes producing different layouts.
Next I looked into the most convoluted card which had five numbers followed by four blanks in the first row and four blanks followed five number in the second. This satisfied the "no blank column" requirement. Now the final row could be arranged however it wanted, though there was one confusing overlap to deal with. 
I considered another convoluted card which had as many columns of three as possible. It turns out it's possible have a maximum of three columns of three which also gives rise to the maximum of six columns of one. It also turns out you can have a maximum of six columns of two. I think the starting point might be to determine the classes of card initially by column combinations and then permute them.
Having gotten all muddled by the layouts, I turned to the numbers. With a single number, it was one of 9, 10 or 11 depending on the column and that was easy. For the two number case, let's say with a column of ten, the first ($i$) was chosen from 9 and the next from $10-i$. It suddenly occurred to me that it was just combinations, phew. How to fit that with the whole business of column choice lost me.
It seems possible that for each possible "set" of columns, there will be a way of determining the distribution of possible first columns and adding in an appropriate factor for the number combinations.
Frankly, it's all getting bigger than my coffee table. 
Any simplifications from higher maths I'm missing?
 A: I would start by listing the ${9 \choose 5} = 126 $ row layouts between blanks and numbers.  Then three loops over those 126 is a total of $126^3=2,000,376$ to find all the legal patterns of blanks.  For each legal pattern, you can calculate the number of ways to fill it with numbers, for example if there are three numbers in the first column, there are ${9 \choose 3}=84$ ways to fill in those numbers.  It shouldn't take a computer long, but I don't know an approach that will work by hand.
A: Much less elegantly than mjqxxxx, I arrived at the count of unnumbered cards by listing them.
Consider each column to have a defining number between 0 (no numbers) and 7 (all three). Subject to the limitation that each row has five numbers and no column has none, they extend from {7,6,6,6,6,1,1,1,1} to {1,1,2,2,4,4,7,7,7}, with a total of 735,210, the same result as above.
My answer for the numbered cards is a little different, though. For the first pattern, there are (in Excel-speak),
=PRODUCT(COMBIN({9,10,10,10,10,10,10,10,11}, {3,2,2,2,2,1,1,1,1})) = 3,788,977,500,000 possibilities.
Summing that result for each row, I get 3,669,688,706,215,250,000 for the count of numbered cards, which agrees to the first 14 digits with the prior result. 
But I'm an Excel guy, not a mathematician.
EDIT: And the difference is the result of running out of precision in a Double, sorry.
