How to calculate probabilities for Gold Lotto?

Question

For my assignment, I am trying to calculate the probabilities of Australian Saturday Gold Lotto (45 balls, 6 winning, 2 supplementary). I need help finding the division calculations to find the division odds (this must use combinations).
gold lotto: div 1 = 6 winning, div 2 = 5 winning numbers + 1 or 2 supplementary, div 3 = 5 winning numbers, div 4 = 4 winning div 5 = 3 winning+ 1 or 2 supplementary, div 6 = 1 or 2 winning numbers + 2 supplementary numbers.
The odds should end up being div 1 = 8145060:1, div 2 = 678755:1, div 3 = 36689:1, div 4 = 733:1 , div 5 = 297:1, div 6 = 144:1
(div 1 & 2 already done)
I have tried several variations of div 1 = 45c7 div 2 = (45c7 / (7c6 x (38c1 - 36c1))) div 3 = (45c7 / (7c6 x 36c1)) div 4 = (45c7 / (7c5 x (38c2 - 36c2))) div 5 = (45c7 / (7c5 x 36c2)) div 6 = (45c7 / (7c4 x 38c3)) div = (45c7 / (7c3 x (38c4 - 36c4)))
(these equations are from oz lotto which is similar to gold lotto)

Answer 1

Strictly speaking, odds against means ratio of unfavorable outcomes:favorable outcomes, whereas what are apparently being computed is total outcomes:favorable outcomes i.e. $\dfrac{1}{p}$.

So I am also using $\dfrac{1}{p}$ while commenting.

Div 1: This is relatively simple,$6$ from winners and none from the rest, thus $\dfrac{\binom{45}6}{\binom66\binom{39}0} = 8145060:1$

Div 2: Not possible, if you have $5$ winning, you can't have $5$ winning and $2$ supplementary.

Div 3 and Div 4: You can do similar to Div 1

Div 5: There are $6$ winning, $2$ supplementary and $37$ others

Winners can be chosen in $\binom63$ ways, and at least one supplementary in $\binom{39}3 -\binom {37}3$ ways,

thus $\dfrac{\binom{45}{6}}{\binom63 \left[\binom{39}3 -\binom{37}3\right]} = 297:1$

You should be able to solve whatever else remains in similar fashion.

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Added for Div 2

If the questioner wants you to ignore the possibility of $5$ winners and $2$ supplementary (which is impossible), and thus reduce the question to $5$ winners, and one supplementary, the answer is $\dfrac{\binom{45}{6}}{\binom65\binom21\binom{37}0} = 678755:1$