# Thread: [SOLVED] Is this right? card problem

1. ## [SOLVED] Is this right? card problem

a deck has 52 cards, 12 face cards and 4 aces are removed, from these 16 cards 4 are chosen, how many combinations are possible that have at least 2 black cards.

I get 154?
What I did was,

there are 8 black cards so,

8c2+8c3+8c4=154
I could swear I did this right.

2. wait I think it's 8c2 times 8c2 ....

Yes it is.

I figured it out....

3. Hello, brentwoodbc!

You're forgetting about the red cards . . .

A deck has 52 cards.
12 face cards and 4 aces are removed.
From these 16 cards, 4 are chosen.
How many combinations are possible that have at least 2 black 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.

4. 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.

Ya I know thank you.