Odds of pulling out 10 specific cards out of a deck

Question

What would the formula be to calculate the odds be of pulling all four 2's, all four 3's and at least 2 of the 4 4's from a standard 52 card deck. No order is necessary as long as these ten cards are pulled first.

A second question is what is the formula or odds of having any three of the four 8's and any 2 of the 4 7's remaining as the last five cards in a standard deck of cards.

Answer 1

Different possible sought after combinations: get any two 4's out of all four of em, so $\binom {4}{2} $

Different possible combinations: Any set of 10 from 52 cards, so $\binom {52}{10}$

$$P = \frac{\binom {4}{2}}{\binom {52}{10}} = \frac6{\frac{52!}{10!42!}}$$

Answer 2

Firstly I'd like to clear up any confusion regarding the term "odds" and how I'll be using it in my answer. If I was to roll a standard fair dice I would have 1 chance in 6 (number of successful outcomes verses the number of all possible outcomes) of rolling a specified number.

If a fair bookmaker (there's an oxymoron) was to take bets on this game he would give odds of 5 to 1, you would (on average) lose 5 times at 1 dollar per time and win 1 time at $5 (the odds the bookmaker gave you ) for every 6 games, leaving you square. Similarly in a standard deck of cards you have 1 chance in 52 of picking the ace of hearts, but the bookmakers odds would be 51 to 1, and so on for various other games of chance. This seeming discrepancy of "1" is only due to the fact that the bookmaker doesn't include your ORIGINAL bet in the odds otherwise it would be the same as the number of chances, either way as the chances increase the discrepancy becomes more insignificant.
Just to be clear, I will be using the number of "CHANCES" as opposes to "ODDS", if you wish to convert, just deduct 1, but in this case with the numbers involved this becomes overwhelmingly insignificant.

Q1
Number of ways 10 cards can be picked from a deck (order unnecessary) = number of ways all four 2's, all four 3's and at least 2 of the 4 4's can be arranged just comes down to how many ways to pick just two of the 4,s (again order excluded)=$\binom {4}{2}$=6
The chances of this happening in the first 10 cards is $\frac{15820024220}{6}\approx2,636,670,703 $

Q2
Number of arrangements (order excluded) of the last 5 cards in a deck = $\binom {52}{5}$=2598960
Number of ways of picking 3 of the 8's = same as the number of ways of excluding 1 of the 8's = 4
Number of ways of picking 2 of the 7's =$\binom {4}{2}$=6
Therefore there are 4x6 = 24 ways of picking the three 8's and two 7's
So the chances of this occurring = $\frac{2598960}{24} =108290$