# In how many different ways can you draw a card (without return) so that your cards have exactly two aces and two red cards?

Here is the question:
In how many different ways can you draw a card (without return) so that your cards have exactly two aces and two red cards? (deck of cards has 52 cards in total with 26 red cards and 26 black cards, 2 of the red cards are aces and 2 of the black cards are aces.)
(a) when drawing 4 cards
(b) when drawing 5 cards
(c) when drawing 6 cards
(d) when drawing 7 cards
*order does matter

My approach to the question is first to find/split possible cases, for example for (a): i can draw 2 red aces and then 2 black not aces, or one red ace, one black ace, one red not ace, one black not ace or 2 black aces and 2red not aces.

Then for solving this i was thinking: $$C(4,2)*2*24*23 + C(4,1)*2*C(3,1)*2*C(2,1)*24^2+C(4,2)*2*24*23$$

Is my approach to the question correct?

• Forgive me but I don't understand one part: Does order matter? Is drawing an ace of spades first and ace of clubs second different than drawing an ace of clubs first and ace of spades second? May 17 at 19:26
• I wrote the question as it is written, I just translated it from another language, from my knowledge,order does matter May 17 at 19:30

Let's do combinations of four cards.

• Two red aces: One way to choose the two red aces. $$24 \cdot 23 / 2$$ ways to choose the black non-aces.
• One red ace, one black ace: Two ways to choose the red ace, and two ways to choose the black ace. $$24$$ ways each to choose the black non-ace and the red non-ace.
• Two black aces: One way to choose the two black aces. $$24 \cdot 23 / 2$$ ways to choose the red non-aces.

So $$(24)(23) + 2(2)(24)(24)$$ combinations.

Multiply everything by $$4!$$ to get the number of permutations since order matters.

Can you take the others from here?

• so basically from what i understand the answer for (a) is $4![2(24)(23)+2(2)(24)(24)]?$ yes i think i can take the other from here, thank you May 18 at 6:42