Six random cards are dealt from a standard deck of 52 playing cards. 
There are 26 red cards and 26 black cards in the deck.  I need to know
the possibility of each possible distribution of colors.  So I need to
know how often all six cards will be black; how often there will be
five black cards and one red and so on.  I also need the same
information for if there are 8, 10, or 12 cards dealt out instead of
6.

For six cards the probabilities are calculated as follows:

6 Blk/0 Red = (26/52)*(25/51)*(24/50)*(23/49)*(22/48)*(21*47)*6!(6!*0!)=0.011309
5 Blk/1 Red = (26/52)*(25/51)*(24/50)*(23/49)*(22/48)*(26*47)*6!(5!*1!)=0.084008
4 Blk/2 Red = (26/52)*(25/51)*(24/50)*(23/49)*(26/48)*(25*47)*6!(4!*2!)=0.238659
3 Blk/3 Red = (26/52)*(25/51)*(24/50)*(26/49)*(25/48)*(24*47)*6!(3!*3!)=0.332048
2 Blk/4 Red = (26/52)*(25/51)*(26/50)*(25/49)*(24/48)*(23*47)*6!(2!*4!)=0.238659
1 Blk/5 Red = (26/52)*(26/51)*(25/50)*(24/49)*(23/48)*(22*47)*6!(1!*5!)=0.084008
0 Blk/6 Red = (26/52)*(25/51)*(24/50)*(23/49)*(22/48)*(21*47)*6!(0!*6!)=0.011309

Total                                                                   1.000000

The same methodology can be used for 8, 10, or 12 cards.

Correction.  I left out a division symbol in the post above.

6 Blk/0Red = (26/52)*(25/51)*(24/50)*(23/49)*(22/48)*(21*47)*6!/(6!*0!)=0.011309
5 Blk/1Red = (26/52)*(25/51)*(24/50)*(23/49)*(22/48)*(26*47)*6!/(5!*1!)=0.084008
4 Blk/2Red = (26/52)*(25/51)*(24/50)*(23/49)*(26/48)*(25*47)*6!/(4!*2!)=0.238659
3 Blk/3Red = (26/52)*(25/51)*(24/50)*(26/49)*(25/48)*(24*47)*6!/(3!*3!)=0.332048
2 Blk/4Red = (26/52)*(25/51)*(26/50)*(25/49)*(24/48)*(23*47)*6!/(2!*4!)=0.238659
1 Blk/5Red = (26/52)*(26/51)*(25/50)*(24/49)*(23/48)*(22*47)*6!/(1!*5!)=0.084008
0 Blk/6Red = (26/52)*(25/51)*(24/50)*(23/49)*(22/48)*(21*47)*6!/(0!*6!)=0.011309

Total                                                                   1.000000

The same methodology can be used for 8, 10, or 12 cards.