How many ways are there to choose from a deck of cards?
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The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.
Choosing 4 from 1 suit (4+1+1+1)
$$4choose1cdot13choose4cdot13choose1^3$$Choosing 2 from 3 suits (2+2+2+1)
$$4choose1cdot13choose2^2cdot13choose1$$Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
$$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$
Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?
combinatorics discrete-mathematics
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0
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The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.
Choosing 4 from 1 suit (4+1+1+1)
$$4choose1cdot13choose4cdot13choose1^3$$Choosing 2 from 3 suits (2+2+2+1)
$$4choose1cdot13choose2^2cdot13choose1$$Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
$$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$
Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?
combinatorics discrete-mathematics
3
First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
â Servaes
Aug 22 at 8:30
5
Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
â Servaes
Aug 22 at 8:34
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.
Choosing 4 from 1 suit (4+1+1+1)
$$4choose1cdot13choose4cdot13choose1^3$$Choosing 2 from 3 suits (2+2+2+1)
$$4choose1cdot13choose2^2cdot13choose1$$Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
$$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$
Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?
combinatorics discrete-mathematics
The prompt is How many ways are there to choose 7 cards from a deck of 52 cards, given that one card from each suit must be chosen?
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us $$4+3-1choose 3 = 20 space ways$$
The other way that I seemed to find on the internet that also makes sense goes like this.
Choosing 4 from 1 suit (4+1+1+1)
$$4choose1cdot13choose4cdot13choose1^3$$Choosing 2 from 3 suits (2+2+2+1)
$$4choose1cdot13choose2^2cdot13choose1$$Choosing 3 from 1 suit and 2 from 2 other (3+2+1+1)
$$4choose1cdot13choose4cdot3choose1cdot13choose2cdot13choose1^2$$
Which also makes sense but gives obviously a different answer than the first method, what am I missing? Which way is the right way?
combinatorics discrete-mathematics
asked Aug 22 at 8:25
Archetype2142
460313
460313
3
First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
â Servaes
Aug 22 at 8:30
5
Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
â Servaes
Aug 22 at 8:34
add a comment |Â
3
First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
â Servaes
Aug 22 at 8:30
5
Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
â Servaes
Aug 22 at 8:34
3
3
First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
â Servaes
Aug 22 at 8:30
First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
â Servaes
Aug 22 at 8:30
5
5
Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
â Servaes
Aug 22 at 8:34
Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
â Servaes
Aug 22 at 8:34
add a comment |Â
3 Answers
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up vote
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The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us
Stars-and-bars counts ways to place identical items into distinct containers. Â Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank. Â It is the wrong way to go.
The alternative method does consider the selection of rank. Â This is the right thing to do.
$7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$. Â We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.
$$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$
add a comment |Â
up vote
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This does not answer your question, but provides a route and is too large for a comment.
Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.
The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.
Then by inclusion/exclusion and symmetry:
$$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$
add a comment |Â
up vote
0
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So we have two ways to solve this question:
$1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
$2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
The second one is easier so:
The number of the cards from each suit can be:
$4/1/1/1$ ,
$3/2/1/1$ or
$2/2/2/1$
IâÂÂm going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
IâÂÂm not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , letâÂÂs suppose we are using the $4/1/1/1$ combination(LetâÂÂs suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
Hope this helps!
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us
Stars-and-bars counts ways to place identical items into distinct containers. Â Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank. Â It is the wrong way to go.
The alternative method does consider the selection of rank. Â This is the right thing to do.
$7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$. Â We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.
$$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$
add a comment |Â
up vote
2
down vote
accepted
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us
Stars-and-bars counts ways to place identical items into distinct containers. Â Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank. Â It is the wrong way to go.
The alternative method does consider the selection of rank. Â This is the right thing to do.
$7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$. Â We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.
$$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us
Stars-and-bars counts ways to place identical items into distinct containers. Â Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank. Â It is the wrong way to go.
The alternative method does consider the selection of rank. Â This is the right thing to do.
$7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$. Â We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.
$$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$
The way I solved this was by imagining we already took 1 card from each suit, now we are left with 3 cards to be picked from 4 suits, this can be done easily using stars and bars giving us
Stars-and-bars counts ways to place identical items into distinct containers. Â Thus it is inapplicable to counting the ways to select cards, which are distinguishable by rank. Â It is the wrong way to go.
The alternative method does consider the selection of rank. Â This is the right thing to do.
$7$ can be partitioned into sums of four terms as $(4+1+1+1)$, $(3+2+1+1)$, and $(2+2+2+1)$. Â We count ways to select suits for each amount, and ways to select those amounts of ranks in the selected suits.
$$binom 4 1,3binom 134^1binom131^3+binom41,1,2binom133^1binom132^1binom131^2+binom 43,1binom132^3binom131^1$$
answered Aug 23 at 0:05
Graham Kemp
80.6k43275
80.6k43275
add a comment |Â
add a comment |Â
up vote
2
down vote
This does not answer your question, but provides a route and is too large for a comment.
Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.
The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.
Then by inclusion/exclusion and symmetry:
$$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$
add a comment |Â
up vote
2
down vote
This does not answer your question, but provides a route and is too large for a comment.
Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.
The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.
Then by inclusion/exclusion and symmetry:
$$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This does not answer your question, but provides a route and is too large for a comment.
Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.
The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.
Then by inclusion/exclusion and symmetry:
$$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$
This does not answer your question, but provides a route and is too large for a comment.
Let $S$ denote the collection of sets of $7$ cards that contain at least one spade. Similarly for $D,H,C$.
The collection of sets of $7$ cards that contain no spades is denoted by $S^complement$.
Then by inclusion/exclusion and symmetry:
$$|S^complementcup D^complementcup H^complementcup C^complement|=binom 41|S^complement|-binom42|S^complementcap D^complement|+binom43|S^complementcap D^complementcap H^complement|$$leading to:$$|Scap Dcap Hcap C|=binom527-binom41binom397+binom42binom267-binom43binom137$$
answered Aug 22 at 9:12
drhab
88k541120
88k541120
add a comment |Â
add a comment |Â
up vote
0
down vote
So we have two ways to solve this question:
$1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
$2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
The second one is easier so:
The number of the cards from each suit can be:
$4/1/1/1$ ,
$3/2/1/1$ or
$2/2/2/1$
IâÂÂm going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
IâÂÂm not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , letâÂÂs suppose we are using the $4/1/1/1$ combination(LetâÂÂs suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
Hope this helps!
add a comment |Â
up vote
0
down vote
So we have two ways to solve this question:
$1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
$2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
The second one is easier so:
The number of the cards from each suit can be:
$4/1/1/1$ ,
$3/2/1/1$ or
$2/2/2/1$
IâÂÂm going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
IâÂÂm not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , letâÂÂs suppose we are using the $4/1/1/1$ combination(LetâÂÂs suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
Hope this helps!
add a comment |Â
up vote
0
down vote
up vote
0
down vote
So we have two ways to solve this question:
$1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
$2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
The second one is easier so:
The number of the cards from each suit can be:
$4/1/1/1$ ,
$3/2/1/1$ or
$2/2/2/1$
IâÂÂm going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
IâÂÂm not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , letâÂÂs suppose we are using the $4/1/1/1$ combination(LetâÂÂs suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
Hope this helps!
So we have two ways to solve this question:
$1.$Find the number of ways that we can choose 7 cards that does not include all the 4 suits and then subtract the answer from the number of all the possible ways that we can choose 7 cards.
$2.$Find the number of ways that we can choose 7 cards from the deck with all the suits included.
The second one is easier so:
The number of the cards from each suit can be:
$4/1/1/1$ ,
$3/2/1/1$ or
$2/2/2/1$
IâÂÂm going to use the same approach as your second solution but please note that your second solution is also wrong because for example , in $2/2/2/1$ , it should be $P(13;11,2) ^ 3$ not $P(13;11,2) ^ 2$.
IâÂÂm not going to give you the answer but please note that your second approach is right you just made a mistake in the solution.
And the reason that your first solution is wrong is because you are just calculating the number of ways you can pick 3 cards from 6 cards which is obviously wrong since , letâÂÂs suppose we are using the $4/1/1/1$ combination(LetâÂÂs suppose we have 4 spades) , the number of ways you can pick 4 spades is alone bigger than your answer.
Hope this helps!
answered Aug 22 at 18:28
Borna Ahmadzade
297
297
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3
First of all, your method does not specify the values of the three last cards to be chosen, only which suit they are. So you are off by a very large factor.
â Servaes
Aug 22 at 8:30
5
Second, by your method you count many ways twice or more. For example, when choosing exactly two hearts (say the king and queen), you could first choose the king when choosing one from each suit, and then the queen when picking the remaining three cards. But you could also first choose the queen, and then the king. The result is the same, but you will count both ways.
â Servaes
Aug 22 at 8:34