The permutation/combination puzzle: What are the maximum number of tickets possible in a game of Tambola?

Question

Tambola is an interesting game which is played by numbers(between 1-90) being called out and players crossing out the numbers on the ticket. Prizes are won on particular patterns.

Facts about the ticket:

There are 3 rows and 9 columns in every Tambola Ticket
There are total 90 (1-90) numbers in Tambola
There are 3 rows and 9 columns in every Tambola Ticket
Every ticket has exactly 15 numbers
Every row contains 5 numbers
A column may have 1, 2 or 3 numbers. (There should be at least 1 number in every column)
The numbers in a column are always ascending in order from top to bottom.
A Ticket cannot have same number more than once
Column 1 on any ticket has numbers between 1-9
Column 2 on any ticket has numbers between 10-19
Column 3 on any ticket has numbers between 20-29
Column 4 on any ticket has numbers between 30-39
Column 5 on any ticket has numbers between 40-49
Column 6 on any ticket has numbers between 50-59
Column 7 on any ticket has numbers between 60-69
Column 8 on any ticket has numbers between 70-79
Column 9 on any ticket has numbers between 80-90

So...What are the maximum number of tickets that can be generated???

Answer 1

Please note that there is a mistake in the question. The conditions

Column 1 on any ticket has numbers between 1-9
Column 2 on any ticket has numbers between 10-19
Column 3 on any ticket has numbers between 20-29
Column 4 on any ticket has numbers between 30-39
Column 5 on any ticket has numbers between 40-49
Column 6 on any ticket has numbers between 50-59
Column 7 on any ticket has numbers between 60-69
Column 8 on any ticket has numbers between 70-79
Column 9 on any ticket has numbers between 80-90

are wrong, not only because they don’t represent the actual game, but also because it would be strange to allow 9 numbers in the first column, 11 in the last, and 10 in the rest. In the following, I solve the problem with the real life constraints:

Column 1 on any ticket has numbers between 1-10
Column 2 on any ticket has numbers between 11-20
Column 3 on any ticket has numbers between 21-30
Column 4 on any ticket has numbers between 31-40
Column 5 on any ticket has numbers between 41-50
Column 6 on any ticket has numbers between 51-60
Column 7 on any ticket has numbers between 61-70
Column 8 on any ticket has numbers between 71-80
Column 9 on any ticket has numbers between 81-90

In each ticket, we have $k$ columns with 3 numbers, $l$ columns with 2 numbers and $m$ columns with 1 number, with $k+l+m=9$ and $2k+l=6$. For each 3-numbered column, we have $10\cdot9\cdot8$ choices; for each 2-numbered column, we have $10\cdot9$, and for each 1-numbered column, simply $10$. They are independent from each other, so we have $10^{k+l+m}9^{k+l}8^{k}$ choices, and we have $$\binom{9}{k,l,m}={9!\over k!l!m!}$$ ways to assign them to columns. To actually place the numbers in the columns they belong to, we have $3^{m}$ choices for the nonempty cells of 1-numbered columns, and $3^{l}$ choices for the empty cells of 2-numbered columns. They are relevant, because the patterns receiving prizes in the real game are row-based (this might not be apparent from the original question) and different empty places generate functionally distinct tickets.

Note that $0\le k\le 3,\,\,\,l=6-2k,\,\,\,m=k+3$, and putting it all together, we get the number of distinct tombola tickets:

$$\sum_{k=0}^{3}\binom{9}{k,l,m}10^{k+l+m}9^{k+l}8^{k}3^{l+m}=\sum_{k=0}^{3}{9!\over k!(6-2k)!(k+3)!}10^{9}9^{6-k}8^{k}3^{9-k} =$$

$$ = 3548382664428000000000 \approx 3.5\cdot10^{21}$$

Note that the game in very popular in Italy and there are several sites which produce tickets for neighborhood festivals, church fundraisers, and similar occasions. I don’t link to them because this could be misunderstood as spamming. Just google “cartelle tombola” to find them. What is relevant here is that many of them (possibly around half) don’t follow the constraint of ordered columns. This would increase the number of possible tickets by a factor of $3!^9$, if we count tickets differing only by a permutation of the entire rows as distinct. However, given the rules of the actual game, they would be functionally identical and having two such tickets simultaneously in the game would be an embarrassment. Therefore, I don’t consider them distinct and to get the number of distinct tombola tickets with unordered columns I multiply the preceding result by $6^8$, to get

$$ 5959920297295899648000000000 \approx 5.9\cdot10^{27}$$

Note: the name of the game is TOM-bola, with the stress on the first syllable. In fact, it is related to the English word tumble.

Answer 2

I think that User3366 wrote the correct rules.

The answer is 6.080.082.602.343.750 The code (in Prolog) to reach the answer is:

sum([], 0).
sum([H|T], Result) :- sum(T, Rest), Result is H + Rest.

:- C1 = [[1,9],[2,36],[3,84]],
C2 = [[1,10],[2,45],[3,120]],
C3 = [[1,10],[2,45],[3,120]],
C4 = [[1,10],[2,45],[3,120]],
C5 = [[1,10],[2,45],[3,120]],
C6 = [[1,10],[2,45],[3,120]],
C7 = [[1,10],[2,45],[3,120]],
C8 = [[1,10],[2,45],[3,120]],
C9 = [[1,11],[2,55],[3,165]],
findall(Z,(
member([X1|Y1],C1),
member([X2|Y2],C2),
member([X3|Y3],C3),
member([X4|Y4],C4),
member([X5|Y5],C5),
member([X6|Y6],C6),
member([X7|Y7],C7),
member([X8|Y8],C8),
member([X9|Y9],C9),
15 is X1+X2+X3+X4+X5+X6+X7+X8+X9,Z is Y1*Y2*Y3*Y4*Y5*Y6*Y7*Y8*Y9),L),
sum(L,CARTONES),
write(CARTONES).

If you read spanish you can find more info here: http://pablopilotti.blogspot.com.ar/2011/02/cartones-de-bingo.html