# COMBINATORY LOGIC: Cards extraction from a deck of 32 cards.

5 cards are extracted simultaneously from a standard deck of 32 cards (8 cards for each of the four suits (hearts, diamonds, spades and clubs): 7,8,9,10, Jack, Queen, King, Ace).

How many different ways can you extract 5 cards containing exactly 3 hearts and exactly 2 kings?

The answer of my book is: 1428 different ways, I am not able to achieve this.

# first case

The two kings are not either of their hearts.

$\binom{3}{2} = 6$ possibilities ==> The hearts cards that remain are EIGHT !! But the king of hearts must be excluded, otherwise the kings extracts become three !!! So are 7 !!!

$$6 \times \binom{7}{3} = 1260$$

# second case

One of the two king of hearts.

6 always possible ==> The hearts cards that remain are SEVEN because it lacks the king of hearts

$$6 \times \binom{7}{2} = 6 \cdot 7 \cdot 6 = 252$$

But $1260 + 252 = 1512 \ne 1428$

What am I doing in the wrong way?

Thank you very much for considering my request.

• Ciao Aurelio, welcome to Maths SE. Try to format your question. There are bits that are quiet unintelligible and bits that read "Google Traduttore per il Business" – Andrea May 23 '16 at 14:30
• Note that $\binom32=3\neq6$. That however is not the only thing that went wrong. – drhab May 23 '16 at 15:04

$$\binom{3}{2}\binom{7}{3}\binom{22}{0}+\binom{1}{1}\binom{3}{1}\binom{7}{2}\binom{21}{1}=3\times35\times1+1\times3\times21\times21=1428$$

Do you see why?

The first term deals with the case that $2$ kings are selected from the $3$ non-heart kings, $3$ hearts from the $7$ non-king hearts and $0$ from the rest.

The second term: $1$ heart-king from $1$ heart-king, $1$ king of $3$ nonheart-kings, $2$ hearts from $7$ nonking-hearts and $1$ from the rest.

• Sounds perfect, but please, a bit of remarks, I am looking forward to click ACCEPTED ANSWER ;-) – Aurelio Severino May 23 '16 at 15:05
• The two non-heart kings, multiplied by 7 heard cards (including heart king), multiplied by combinations of 22 on zero this means 1. Plus: the heart king multiplied by the combinations of 3 non-heart kings multiplied by comb(7,2) heard cards multiplied by comb(21,1) for one remaining card... Is the remark right? – Aurelio Severino May 23 '16 at 15:09
• Eh.. I think so, but check yourself on base of my edit. – drhab May 23 '16 at 15:11
• Answer accepted, compliments and many thanks for having helped me! ;) I am not the cards' king!! :-D – Aurelio Severino May 23 '16 at 15:12
• Glad to help. You are welcome. – drhab May 23 '16 at 15:13

Case 1: There are not $6$ ways to choose the non-heart kings, rather $3$, as $\binom{3}{2}=\frac{3!}{2!1!}=3$.

Case 2: Again, $6$ is the wrong number to use. Once you have chosen the non-king hearts and the king of hearts, you need to choose two more cards: one which is a king other than the king of hearts (how many ways are there to choose this card?), and one which is a non-king and non-heart (how many ways are there to choose this card?).

You will get $1428$ if you use the correct numbers.

• could you show me the calculus please? – Aurelio Severino May 23 '16 at 14:43