If I was at a Casino and lost over $20,000 in a slot machine before hitting a $10,000 jackpot, will I still have to fill out a tax form and declare that as income even thought I really lost $10,000?

There are two ways to handle slot winnings/losses. The regular way is that all winnings (whether you receive a W-2G or not) are reported on your 1040 as other income, and your losses that you can substantiate are deducted on your Schedule A (assuming that you itemize, etc.). Under that method, yes, you would report the $10K jackpot (plus any other winnings that came out of the machine(s)), and you would deduct your losses up to the amount of your reported winnings as an itemized deduction.The other way to do it is the technically correct way — and that is to net your slot results on a session basis. That is, from the time that you pull your first lever (or push your first spin button) of the day until you stop playing the slots, other than short breaks — but not beyond midnight — you total your net winnings/losses, and that’s your winnings or losses for the session that are reportable on your tax return. That method, for example, enables you to take advantage of losses even if you don’t itemize deductions. Now, that’s not going to match up with any W-2G’s that you get, because the casinos don’t report on that basis (they don’t even report on the basis of a midnight-to-midnight day). But the IRS has ways in which you can indicate that on your return.So the answer to your question is, maybe. It depends on (i) when did you lose the $20K and win the $10K?, (ii) what kind of records do you have (a players’ card statement would be great), and (iii) what else did you win gambling?

In Bingo, where you have 15 out of 90 numbers, a grid 3 by 9, you must have a least 1 number from every group, 1 to 9, 10 to 19, 20 to 29 etc, and no more than 3 numbers from each. What is the math to work out how many permutations there are ?

Please explain the role of the 3x8 grid. Normal bingo cards/grids are 5x5. If you are thinking that each column contains a unique group (using your terminology), then you need a 3x9 group.EDIT: I thought a bit about your question this morning and think I have a guess at what you were trying to ask. Let me see if this is correct.You have a total of 90 balls each with a unique number. You also have a grid 9 rows and 3 columns. You draw 15 balls at random and place them on the grid based on which group (1–9, 10–10, 20–29, … , 80–90). You want to know how may ways the grid can be filled in where there is at least 1 ball in each row but no more than 3. Because you used the word permutation, I’m going to assume that you do care about the order of the balls in each row, i.e. 20,26,23 is not the same as 23,26,20.Before answering this, I’m going to suggest that we redefine the groups as 1–10, 2–20, …, 81–90. Otherwise, there is a nasty asymmetry occurring at the endpoints. The first row only has 9 possible balls to fill it. The last row has 11. All other rows have 10. This tweak will make the math a whole lot easier.So, on to the solution…The first step is to figure out how many ways you can partition the 15 balls into 9 groups of 1, 2, or 3 balls. We know that each of the 9 groups must contain at least 1 ball. This means we must distribute the remaining 6 balls into groups of 1 or 2 balls each. There are 4 possible ways to do this: {2,2,2}, {2,2,1,1},{2,1,1,1,1},{1,1,1,1,1,1}. These correspond to 4 partitions of the 15 balls into 9 groups of 1, 2, or 3 balls: {3,3,3,1,1,1,1,1,1}, {3,3,2,2,1,1,1,1,1},{3,2,2,2,2,1,1,1,1},{2,2,2,2,2,2,1,1,1}.The second step is to figure out how to distribute each of these partitions to the actual rows in the grid. But, we must address each on its own.The first partition contains three 3’s and six 1’s. There are C(9,3) = (9*8*7)/(3*2*1) = 84 ways to pick the three rows which will contain the 3’s. There is only 1 way to pick the six rows to contain the 1’s. This gives a total of 84 patterns associated with this partition.The second partition contains two 3’s, two 2’s and five 1’s. There are C(9,2) = (9*8)/(2*1) = 36 ways to pick the 2 rows which will contain the 3’s. There are C(7,2) = (7*6)/(2*1) = 21 ways to pick the 2 rows will will contain the 2’s. Again, there is only 1 way to pick the 5 rows to contain the 1’s. This gives a total of 36*21 = 756 patterns associated with this partition.The third partition contains one 3, four 2’s and four 1’s. There are C(9,1) = 9 ways to pick the row which will contain the 3. There are C(8,4)=(8*7*6*5)/(4*3*2*1) = 70 ways to pick the four rows will will contain the 2’s. Again, there is only one way to pick the rows to contain the 1’s. This gives a total of 9*70 = 630 patterns associated with this partition.The fourth partition contains six 2’s and 3 1’s. There are C(9,6) = (9*8*7*6*5*4)/(6*5*4*3*2*1) = 84 ways to pick the rows which will contain the 2’s. Yet again, there is only one way to pick the rows to contain the 1’s. This gives a total of 84 patterns associated with this partition.The next step is to figure out how to distribute the balls into each of these rows.Any row with a single ball has 10 possible ways to pick that ball.Any row with two balls has C(10,2) = (10*9)/(2*1) = 45 ways to pick them if order of balls in the row doesn’t matter or P(10,2) = 10*9 = 90 ways if the order does matter.Any row with three balls has C(10,3) = (10*9*8)/(3*2*1) = 120 ways to pick them if order doesn’t matter or P(10,3) = 10*9*8 = 720 if the order does matter.The first partition pattern has therefore, 120^3 * 10^6 = 1,728,000,000,000 ways to distribute for each pattern where order doesn’t matter or 720^3 * 10^6 = 373,248,000,000,000 where it does. This yields totals of 145,152,000,000,000 for the 84 patterns where order doesn’t matter or 31,352,832,000,000,000 where it does.The second partition has 120^2*45^2*10^5 = 291,600,000,000 ways to distribute for each pattern where order doesn’t matter or 720^2 * 90^2 * 10^5 = 419,904,000,000,000 where it does. This yields totals of 220,449,600,000,000 for the 756 patterns where order doesn’t matter or 317,447,424,000,000,000 where it does.The third partition has 120*45^4*10^4 = 4,920,750,000,000 ways to distribute for each pattern where order doesn’t matter or 720 * 90^4 * 10^4 = 472,392,000,000,000 where it does. This yields totals of 31,000,725,000,000,000 for the 630 patterns where order doesn’t matter or 297,606,960,000,000,000 where it does.The third partition has 45^6*10^3 = 8,303,765,625,000 ways to distribute for each pattern where order doesn’t matter or 90^6 * 10^3 = 531,441,000,000,000 where it does. This yields totals of 697,516,312,500,000 for the 84 patterns where order doesn’t matter or 44,641,044,000,000,000 where it does.I will let you do the additions yourself. But as another Quoran answered, a lot!

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well state officials say illinois drivers age 75 and older will get an extra year to renew their driver's licenses because of the coronavirus pandemic the new expiration date will be a year from the current 2020 date the secretary of state's office says that means that seniors don't need to visit a facility to renew their driver's license until shortly before their birthday in 2021

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If I was at a Casino and lost over $20,000 in a slot machine before hitting a $10,000 jackpot, will I still have to fill out a tax form and declare that as income even thought I really lost $10,000?
There are two ways to handle slot winnings/losses. The “regular” way is that all winnings (whether you receive a W-2G or not) are reported on your 1040 as “other income”, and your losses that you can substantiate are deducted on your Schedule A (assuming that you itemize, etc.). Under that method, yes, you would report the $10K jackpot (plus any other winnings that came out of the machine(s)), and you would deduct your losses up to the amount of your reported winnings as an itemized deduction.The other way to do it is the technically correct way — and that is to net your slot results on a “session” basis. That is, from the time that you pull your first lever (or push your first “spin” button”) of the day until you stop playing the slots, other than short breaks — but not beyond midnight — you total your net winnings/losses, and that’s your winnings or losses for the “session” that are reportable on your tax return. That method, for example, enables you to take advantage of losses even if you don’t itemize deductions. Now, that’s not going to match up with any W-2G’s that you get, because the casinos don’t report on that basis (they don’t even report on the basis of a midnight-to-midnight day). But the IRS has ways in which you can indicate that on your return.So the answer to your question is, maybe. It depends on (i) when did you lose the $20K and win the $10K?, (ii) what kind of records do you have (a “players’ card” statement would be great), and (iii) what else did you win gambling?
In Bingo, where you have 15 out of 90 numbers, a grid 3 by 9, you must have a least 1 number from every group, 1 to 9, 10 to 19, 20 to 29 etc, and no more than 3 numbers from each. What is the math to work out how many permutations there are ?
Please explain the role of the 3x8 grid. Normal bingo cards/grids are 5x5. If you are thinking that each column contains a unique group (using your terminology), then you need a 3x9 group.EDIT: I thought a bit about your question this morning and think I have a guess at what you were trying to ask. Let me see if this is correct.You have a total of 90 balls each with a unique number. You also have a grid 9 rows and 3 columns. You draw 15 balls at random and place them on the grid based on which group (1–9, 10–10, 20–29, … , 80–90). You want to know how may ways the grid can be filled in where there is at least 1 ball in each row but no more than 3. Because you used the word permutation, I’m going to assume that you do care about the order of the balls in each row, i.e. 20,26,23 is not the same as 23,26,20.Before answering this, I’m going to suggest that we redefine the groups as 1–10, 2–20, …, 81–90. Otherwise, there is a nasty asymmetry occurring at the endpoints. The first row only has 9 possible balls to fill it. The last row has 11. All other rows have 10. This tweak will make the math a whole lot easier.So, on to the solution…The first step is to figure out how many ways you can partition the 15 balls into 9 groups of 1, 2, or 3 balls. We know that each of the 9 groups must contain at least 1 ball. This means we must distribute the remaining 6 balls into groups of 1 or 2 balls each. There are 4 possible ways to do this: {2,2,2}, {2,2,1,1},{2,1,1,1,1},{1,1,1,1,1,1}. These correspond to 4 partitions of the 15 balls into 9 groups of 1, 2, or 3 balls: {3,3,3,1,1,1,1,1,1}, {3,3,2,2,1,1,1,1,1},{3,2,2,2,2,1,1,1,1},{2,2,2,2,2,2,1,1,1}.The second step is to figure out how to distribute each of these partitions to the actual rows in the grid. But, we must address each on its own.The first partition contains three 3’s and six 1’s. There are C(9,3) = (9*8*7)/(3*2*1) = 84 ways to pick the three rows which will contain the 3’s. There is only 1 way to pick the six rows to contain the 1’s. This gives a total of 84 patterns associated with this partition.The second partition contains two 3’s, two 2’s and five 1’s. There are C(9,2) = (9*8)/(2*1) = 36 ways to pick the 2 rows which will contain the 3’s. There are C(7,2) = (7*6)/(2*1) = 21 ways to pick the 2 rows will will contain the 2’s. Again, there is only 1 way to pick the 5 rows to contain the 1’s. This gives a total of 36*21 = 756 patterns associated with this partition.The third partition contains one 3, four 2’s and four 1’s. There are C(9,1) = 9 ways to pick the row which will contain the 3. There are C(8,4)=(8*7*6*5)/(4*3*2*1) = 70 ways to pick the four rows will will contain the 2’s. Again, there is only one way to pick the rows to contain the 1’s. This gives a total of 9*70 = 630 patterns associated with this partition.The fourth partition contains six 2’s and 3 1’s. There are C(9,6) = (9*8*7*6*5*4)/(6*5*4*3*2*1) = 84 ways to pick the rows which will contain the 2’s. Yet again, there is only one way to pick the rows to contain the 1’s. This gives a total of 84 patterns associated with this partition.The next step is to figure out how to distribute the balls into each of these rows.Any row with a single ball has 10 possible ways to pick that ball.Any row with two balls has C(10,2) = (10*9)/(2*1) = 45 ways to pick them if order of balls in the row doesn’t matter or P(10,2) = 10*9 = 90 ways if the order does matter.Any row with three balls has C(10,3) = (10*9*8)/(3*2*1) = 120 ways to pick them if order doesn’t matter or P(10,3) = 10*9*8 = 720 if the order does matter.The first partition pattern has therefore, 120^3 * 10^6 = 1,728,000,000,000 ways to distribute for each pattern where order doesn’t matter or 720^3 * 10^6 = 373,248,000,000,000 where it does. This yields totals of 145,152,000,000,000 for the 84 patterns where order doesn’t matter or 31,352,832,000,000,000 where it does.The second partition has 120^2*45^2*10^5 = 291,600,000,000 ways to distribute for each pattern where order doesn’t matter or 720^2 * 90^2 * 10^5 = 419,904,000,000,000 where it does. This yields totals of 220,449,600,000,000 for the 756 patterns where order doesn’t matter or 317,447,424,000,000,000 where it does.The third partition has 120*45^4*10^4 = 4,920,750,000,000 ways to distribute for each pattern where order doesn’t matter or 720 * 90^4 * 10^4 = 472,392,000,000,000 where it does. This yields totals of 31,000,725,000,000,000 for the 630 patterns where order doesn’t matter or 297,606,960,000,000,000 where it does.The third partition has 45^6*10^3 = 8,303,765,625,000 ways to distribute for each pattern where order doesn’t matter or 90^6 * 10^3 = 531,441,000,000,000 where it does. This yields totals of 697,516,312,500,000 for the 84 patterns where order doesn’t matter or 44,641,044,000,000,000 where it does.I will let you do the additions yourself. But as another Quoran answered, “a lot!”

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