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 Nov 15 at 8:22
Robert Z
90.4k1056128
asked Nov 15 at 7:35
Konno Monno
31
add a comment |
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 Nov 15 at 8:22
Robert Z
90.4k1056128
asked Nov 15 at 7:35
Konno Monno
31
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
Nov 15 at 7:48
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 Nov 15 at 8:22
Robert Z
90.4k1056128
asked Nov 15 at 7:35
Konno Monno
31
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 Nov 15 at 8:22
Robert Z
90.4k1056128
asked Nov 15 at 7:35
Konno Monno
31
edited Nov 15 at 8:22
Robert Z
90.4k1056128
asked Nov 15 at 7:35
Konno Monno
31
edited Nov 15 at 8:22
Robert Z
90.4k1056128
edited Nov 15 at 8:22
Robert Z
90.4k1056128
edited Nov 15 at 8:22
Robert Z
90.4k1056128
90.4k1056128
asked Nov 15 at 7:35
Konno Monno
31
asked Nov 15 at 7:35
Konno Monno
31
asked Nov 15 at 7:35
Konno Monno
31
31
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
Nov 15 at 7:48
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
Nov 15 at 7:48
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
Nov 15 at 7:48
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
Nov 15 at 7:48
add a comment |
1 Answer
1
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 Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
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
Nov 15 at 7:53
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 Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
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
Nov 15 at 7:53
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 Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
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
Nov 15 at 7:53
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 Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
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 Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
edited Nov 15 at 7:51
answered Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
answered Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
answered Nov 15 at 7:44
Siong Thye Goh
93.8k1462114
93.8k1462114
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
Nov 15 at 7:53
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
Nov 15 at 7:53
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
Nov 15 at 7:53
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
Nov 15 at 7:53
add a comment |
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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
Nov 15 at 7:48