# How much money would it cost to “buy out” every possible number in the 6-digit Lottery?

So, the constraints are as follows:

• First 5 numbers are digits 1 ; 69
• The 6th number is a number 1 - 26
• Order does not matter
• Numbers do not repeat
• Tickets cost $2 each My reasoning is as follows: for the first 5 digits, there are clearly $$69 \times 68 \times 67 \times 66 \times 65$$ possibilities. For the 6th digit, there are 26 possible choices, but we must account for the fact that the previous digits could've been within the range of $$[1, 26]$$ (or we end up overcounting). This is where I'm a little less certain: clearly I can't just multiply the previous number by 26. The "worst" case is where all 5 previous numbers were also in the range of $$[1, 26]$$; the "best" case is where all of them were in the range $$[27, 69]$$. My thought process was to just multiply my previous number by $$21$$ (because $$26 - 5 = 21$$), but I think that this may be undercounting because there are $$43 \times 42 \times 41 \times 40 \times 39$$ possible numbers that they could take that are greater than 26. It's easy to account for the$2 tickets because if I want to calculate how much money it would take to buy enough tickets to guarantee a win, I'd just have to multiply my previous result by 2.

So, my proposed solution is that it would cost $$69 \times 68 \times 67 \times 66 \times 65 \times 21 \times 2$$ dollars to buy enough Lottery tickets to guarantee that you would win.

Can someone help me with the number of possible values for the final digits? Is $$21$$ correct, or are there more?

• Try choosing the 6th number first. – FredH Mar 28 at 18:11

$$\dbinom{26}{1}\dbinom{68}{5}$$