Possible ways to choose 3 Kings and 2 other non-paired cards?
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From a standard deck (52 cards, 4 suits) of playing cards, how many ways is it possible to be dealt three kings as a three of a kind? So the two other cards cannot be able to form a pair nor can they be another king as that would make it a four of a kind.
This is the answer I came up with, however, I feel like it is wrong. There is a possibility that this expression allows me to pick any three of a kind.
$$[C(13,1)cdot C(4,3)] cdot [C(12,2)cdot C(4,1) cdot C(4,1)]$$
To prevent such occurrences, would I have to divide the expression with "13!"? Or would I have to rewrite the expression in a completely new way? I might be overthinking this but I still feel like I am wrong.
combinatorics
edited 8 hours ago
Robert Z
89.3k1056128
asked 9 hours ago
Konno Monno
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Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
favorite
From a standard deck (52 cards, 4 suits) of playing cards, how many ways is it possible to be dealt three kings as a three of a kind? So the two other cards cannot be able to form a pair nor can they be another king as that would make it a four of a kind.
This is the answer I came up with, however, I feel like it is wrong. There is a possibility that this expression allows me to pick any three of a kind.
$$[C(13,1)cdot C(4,3)] cdot [C(12,2)cdot C(4,1) cdot C(4,1)]$$
To prevent such occurrences, would I have to divide the expression with "13!"? Or would I have to rewrite the expression in a completely new way? I might be overthinking this but I still feel like I am wrong.
combinatorics
edited 8 hours ago
Robert Z
89.3k1056128
asked 9 hours ago
Konno Monno
31
New contributor
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
9 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From a standard deck (52 cards, 4 suits) of playing cards, how many ways is it possible to be dealt three kings as a three of a kind? So the two other cards cannot be able to form a pair nor can they be another king as that would make it a four of a kind.
This is the answer I came up with, however, I feel like it is wrong. There is a possibility that this expression allows me to pick any three of a kind.
$$[C(13,1)cdot C(4,3)] cdot [C(12,2)cdot C(4,1) cdot C(4,1)]$$
To prevent such occurrences, would I have to divide the expression with "13!"? Or would I have to rewrite the expression in a completely new way? I might be overthinking this but I still feel like I am wrong.
combinatorics
edited 8 hours ago
Robert Z
89.3k1056128
asked 9 hours ago
Konno Monno
31
New contributor
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
From a standard deck (52 cards, 4 suits) of playing cards, how many ways is it possible to be dealt three kings as a three of a kind? So the two other cards cannot be able to form a pair nor can they be another king as that would make it a four of a kind.
This is the answer I came up with, however, I feel like it is wrong. There is a possibility that this expression allows me to pick any three of a kind.
$$[C(13,1)cdot C(4,3)] cdot [C(12,2)cdot C(4,1) cdot C(4,1)]$$
To prevent such occurrences, would I have to divide the expression with "13!"? Or would I have to rewrite the expression in a completely new way? I might be overthinking this but I still feel like I am wrong.
combinatorics
combinatorics
edited 8 hours ago
Robert Z
89.3k1056128
asked 9 hours ago
Konno Monno
31
New contributor
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Robert Z
89.3k1056128
asked 9 hours ago
Konno Monno
31
New contributor
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Robert Z
89.3k1056128
edited 8 hours ago
Robert Z
89.3k1056128
edited 8 hours ago
Robert Z
89.3k1056128
89.3k1056128
asked 9 hours ago
Konno Monno
31
New contributor
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 9 hours ago
Konno Monno
31
asked 9 hours ago
Konno Monno
31
31
New contributor
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Konno Monno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
9 hours ago
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
9 hours ago
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
9 hours ago
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
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accepted
First handle the king:$binom{4}3$
and then choose $2$ cards out of the other denomination:$binom{12}2$.
Choose the suit for the two cards: $4^2$.
Hence overall, $$binom{4}3binom{12}24^2=4^3cdot frac{12(11)}{2!}=(66)(64)=4224$$
Remark: There is no need to have $binom{13}1$ since we already know we want that to be the king.
answered 9 hours ago
Siong Thye Goh
92.3k1461114
I see, since it is already established that I have to pick a king there is no need for C(13,1) as this would only make me choose 1 of the 13 possible ranks. Thank you very much!
– Konno Monno
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
First handle the king:$binom{4}3$
and then choose $2$ cards out of the other denomination:$binom{12}2$.
Choose the suit for the two cards: $4^2$.
Hence overall, $$binom{4}3binom{12}24^2=4^3cdot frac{12(11)}{2!}=(66)(64)=4224$$
Remark: There is no need to have $binom{13}1$ since we already know we want that to be the king.
answered 9 hours ago
Siong Thye Goh
92.3k1461114
I see, since it is already established that I have to pick a king there is no need for C(13,1) as this would only make me choose 1 of the 13 possible ranks. Thank you very much!
– Konno Monno
9 hours ago
add a comment |
up vote
0
down vote
accepted
First handle the king:$binom{4}3$
and then choose $2$ cards out of the other denomination:$binom{12}2$.
Choose the suit for the two cards: $4^2$.
Hence overall, $$binom{4}3binom{12}24^2=4^3cdot frac{12(11)}{2!}=(66)(64)=4224$$
Remark: There is no need to have $binom{13}1$ since we already know we want that to be the king.
answered 9 hours ago
Siong Thye Goh
92.3k1461114
I see, since it is already established that I have to pick a king there is no need for C(13,1) as this would only make me choose 1 of the 13 possible ranks. Thank you very much!
– Konno Monno
9 hours ago
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
First handle the king:$binom{4}3$
and then choose $2$ cards out of the other denomination:$binom{12}2$.
Choose the suit for the two cards: $4^2$.
Hence overall, $$binom{4}3binom{12}24^2=4^3cdot frac{12(11)}{2!}=(66)(64)=4224$$
Remark: There is no need to have $binom{13}1$ since we already know we want that to be the king.
answered 9 hours ago
Siong Thye Goh
92.3k1461114
First handle the king:$binom{4}3$
and then choose $2$ cards out of the other denomination:$binom{12}2$.
Choose the suit for the two cards: $4^2$.
Hence overall, $$binom{4}3binom{12}24^2=4^3cdot frac{12(11)}{2!}=(66)(64)=4224$$
Remark: There is no need to have $binom{13}1$ since we already know we want that to be the king.
answered 9 hours ago
Siong Thye Goh
92.3k1461114
edited 9 hours ago
answered 9 hours ago
Siong Thye Goh
92.3k1461114
answered 9 hours ago
Siong Thye Goh
92.3k1461114
answered 9 hours ago
Siong Thye Goh
92.3k1461114
92.3k1461114
I see, since it is already established that I have to pick a king there is no need for C(13,1) as this would only make me choose 1 of the 13 possible ranks. Thank you very much!
– Konno Monno
9 hours ago
add a comment |
I see, since it is already established that I have to pick a king there is no need for C(13,1) as this would only make me choose 1 of the 13 possible ranks. Thank you very much!
– Konno Monno
9 hours ago
I see, since it is already established that I have to pick a king there is no need for C(13,1) as this would only make me choose 1 of the 13 possible ranks. Thank you very much!
– Konno Monno
9 hours ago
I see, since it is already established that I have to pick a king there is no need for C(13,1) as this would only make me choose 1 of the 13 possible ranks. Thank you very much!
– Konno Monno
9 hours ago
add a comment |
Konno Monno is a new contributor. Be nice, and check out our Code of Conduct.
Konno Monno is a new contributor. Be nice, and check out our Code of Conduct.
Konno Monno is a new contributor. Be nice, and check out our Code of Conduct.
Konno Monno is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
9 hours ago