Originally Posted by

**Soroban** Hello, brentwoodbc!

You're forgetting about the red cards . . .

To get "at least 2 Blacks", we want: .(2 Black, 2 Red) or (3 Black, 1 Red) or (4 Black).

. . There are: .$\displaystyle (_8C_2)(_8C_2) = 784$ ways to get 2 Black, 2 Red.

. . There are: .$\displaystyle (_8C_3)(_8C_1) = 448$ ways to get 3 Black, 1 Red.

. . There are: .$\displaystyle _8C_4 = 70$ ways to get 4 Black.

Therefore, there are: .$\displaystyle 784 + 448 + 70 \:=\:1302$ ways to get at least 2 Blacks.