How many 5-card hands from a standard 52-card deck contain exactly 1 king and exactly 1 heart?
$begingroup$
Is my solution correct?
There are two cases:
Case 1: (the chosen heart is not a king) So there are $3 choose 1 $ ways to choose the king, and $12 choose 1 $ ways to choose the heart, and then $52-12-3 choose 3$ ways to choose rest of the three cards.
Case 2: (the chosen heart is a king) So there are $1 choose 1$ way to choose the king (heart) card, and then since rest of the 4 cards cannot be a king nor heart, there are $52-1-12-3 choose 4$ ways to choose the rest.
Then adding these two cases together we get the result.
combinatorics combinations card-games
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
asked Dec 9 '18 at 20:56
TorichanTorichan
242
$endgroup$
add a comment |
$begingroup$
Is my solution correct?
There are two cases:
Case 1: (the chosen heart is not a king) So there are $3 choose 1 $ ways to choose the king, and $12 choose 1 $ ways to choose the heart, and then $52-12-3 choose 3$ ways to choose rest of the three cards.
Case 2: (the chosen heart is a king) So there are $1 choose 1$ way to choose the king (heart) card, and then since rest of the 4 cards cannot be a king nor heart, there are $52-1-12-3 choose 4$ ways to choose the rest.
Then adding these two cases together we get the result.
combinatorics combinations card-games
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
asked Dec 9 '18 at 20:56
TorichanTorichan
242
$endgroup$
add a comment |
$begingroup$
Is my solution correct?
There are two cases:
Case 1: (the chosen heart is not a king) So there are $3 choose 1 $ ways to choose the king, and $12 choose 1 $ ways to choose the heart, and then $52-12-3 choose 3$ ways to choose rest of the three cards.
Case 2: (the chosen heart is a king) So there are $1 choose 1$ way to choose the king (heart) card, and then since rest of the 4 cards cannot be a king nor heart, there are $52-1-12-3 choose 4$ ways to choose the rest.
Then adding these two cases together we get the result.
combinatorics combinations card-games
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
asked Dec 9 '18 at 20:56
TorichanTorichan
242
$endgroup$
Is my solution correct?
There are two cases:
Case 1: (the chosen heart is not a king) So there are $3 choose 1 $ ways to choose the king, and $12 choose 1 $ ways to choose the heart, and then $52-12-3 choose 3$ ways to choose rest of the three cards.
Case 2: (the chosen heart is a king) So there are $1 choose 1$ way to choose the king (heart) card, and then since rest of the 4 cards cannot be a king nor heart, there are $52-1-12-3 choose 4$ ways to choose the rest.
Then adding these two cases together we get the result.
combinatorics combinations card-games
combinatorics combinations card-games
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
asked Dec 9 '18 at 20:56
TorichanTorichan
242
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
asked Dec 9 '18 at 20:56
TorichanTorichan
242
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
edited Dec 9 '18 at 21:04
N. F. Taussig
44k93356
44k93356
asked Dec 9 '18 at 20:56
TorichanTorichan
242
asked Dec 9 '18 at 20:56
TorichanTorichan
242
asked Dec 9 '18 at 20:56
TorichanTorichan
242
242
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Almost.
In the first case, once you select a king and a heart that is not the king of hearts, there are $16$ cards from which you cannot select the other three cards: the four kings and the hearts, as $13 + 4 - 1 = 16$. Therefore, you should have
$$binom{3}{1}binom{12}{1}binom{52 - 13 - 4 + 1}{3}$$
Your second case is fine.
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Almost.
In the first case, once you select a king and a heart that is not the king of hearts, there are $16$ cards from which you cannot select the other three cards: the four kings and the hearts, as $13 + 4 - 1 = 16$. Therefore, you should have
$$binom{3}{1}binom{12}{1}binom{52 - 13 - 4 + 1}{3}$$
Your second case is fine.
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
add a comment |
$begingroup$
Almost.
In the first case, once you select a king and a heart that is not the king of hearts, there are $16$ cards from which you cannot select the other three cards: the four kings and the hearts, as $13 + 4 - 1 = 16$. Therefore, you should have
$$binom{3}{1}binom{12}{1}binom{52 - 13 - 4 + 1}{3}$$
Your second case is fine.
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
add a comment |
$begingroup$
Almost.
In the first case, once you select a king and a heart that is not the king of hearts, there are $16$ cards from which you cannot select the other three cards: the four kings and the hearts, as $13 + 4 - 1 = 16$. Therefore, you should have
$$binom{3}{1}binom{12}{1}binom{52 - 13 - 4 + 1}{3}$$
Your second case is fine.
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
$endgroup$
Almost.
In the first case, once you select a king and a heart that is not the king of hearts, there are $16$ cards from which you cannot select the other three cards: the four kings and the hearts, as $13 + 4 - 1 = 16$. Therefore, you should have
$$binom{3}{1}binom{12}{1}binom{52 - 13 - 4 + 1}{3}$$
Your second case is fine.
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
answered Dec 9 '18 at 21:03
N. F. TaussigN. F. Taussig
44k93356
44k93356
add a comment |
add a comment |
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