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 yesterday
Robert Z
89.4k1056128
asked yesterday
Konno Monno
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up vote
0
down vote
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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 yesterday
Robert Z
89.4k1056128
asked yesterday
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
yesterday
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 yesterday
Robert Z
89.4k1056128
asked yesterday
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 yesterday
Robert Z
89.4k1056128
asked yesterday
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 yesterday
Robert Z
89.4k1056128
asked yesterday
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 yesterday
Robert Z
89.4k1056128
edited yesterday
Robert Z
89.4k1056128
edited yesterday
Robert Z
89.4k1056128
89.4k1056128
asked yesterday
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 yesterday
Konno Monno
31
asked yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday
add a comment |
1 Answer
1
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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 yesterday
Siong Thye Goh
92.4k1461114
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
yesterday
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 yesterday
Siong Thye Goh
92.4k1461114
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
yesterday
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 yesterday
Siong Thye Goh
92.4k1461114
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
yesterday
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 yesterday
Siong Thye Goh
92.4k1461114
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 yesterday
Siong Thye Goh
92.4k1461114
edited yesterday
answered yesterday
Siong Thye Goh
92.4k1461114
answered yesterday
Siong Thye Goh
92.4k1461114
answered yesterday
Siong Thye Goh
92.4k1461114
92.4k1461114
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
yesterday
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
yesterday
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
yesterday
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
yesterday
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
yesterday