There are C(6,r) ways to choose r of the winning numbers from the six. There are 34 incorrect numbers. We want to choose r of these. There are C(34,r) ways to select these. Therefore there are C(6,r) * C(34,6-r) winning combinations for each selection. But since there are TWO selections then there are 2 * C(6,r) * C(34,6-r) total winning combinations in all.
Two distinct possiblities case 1: All of the numbers in common (1..5) are correct. case 2: Not all of 1..5 are correct
1..5 are correct. Only the last number has to be selected. This is a non winning number. It does not contain [1..5]. It also does not contain [6,7] because then this would make for a 6-win. So 40-7 leaves 33 possibilities. However 1-2-3-4-5-6 produces a 5-win for selection one and 1-2-3-4-5-7 produces a 5-win for selection two. Though they also produce a 6-win for the other selection. So we are force to included these two combinations as well. Thus there are 33+2 = 35 5-win combinations.
One (only one) of 1..5 is incorrect. There are 5 ways for this to happen. Four numbers have been selected at this point. Five less the incorrect number. Seeing that exactly five numbers must be correct then 6 or 7 MUST BE CORRECT. They may both be correct.
case 2a: 6 or 7 have come in (but NOT BOTH). For the fifth number there are 2 possible combinations. 6 or 7. And there are 33 possibilities for the remaining wrong number. Note: NOT 34 as we exclude both 6 and 7 from occurring as the other number. Thus 5 * 2 * 33 = 330 winning possibilities.
case 2b: Both 6 and 7 have come in. Both selection 1 and 2 have won. There are only have combinations that have both 6 and 7 in them. These are:
1-2-3-4-6-7
1-2-3-5-6-7
1-2-4-5-6-7
1-3-4-5-6-7
2-3-4-5-6-7
case 1 + case 2a + case 2b = 35 + 330 + 5 = 370 possible winning combinations.
If both selections one and two contained entirely different numbers then the total winning combinations would be 204 + 204 = 408 winning combinations. However they have winning combinations in common.
case 1: Numbers 1..5 come in. Neither 6 nor 7 come in either so there are 40 - 5 - 2 = 33 possibilities for the remaining number. There are 33 winning combinations in common in this case.
case 2: One of the numbers 1..5 does not come in but instead both 6 and 7 come in. There are 5 possible combinations and these are listed above.
case 1 + case 2 = 33 + 5 = 38 winning combinations in common. Because we have counted these twice we subtract 38 of these. 408 - 38 = 370 possible combinations.
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