# 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" May 23, 2016 at 14:30
• Note that $\binom32=3\neq6$. That however is not the only thing that went wrong. May 23, 2016 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 ;-) May 23, 2016 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? May 23, 2016 at 15:09
• Eh.. I think so, but check yourself on base of my edit. May 23, 2016 at 15:11
• Answer accepted, compliments and many thanks for having helped me! ;) I am not the cards' king!! :-D May 23, 2016 at 15:12
• Glad to help. You are welcome. May 23, 2016 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? May 23, 2016 at 14:43