How many combinations are there to pull at least 1 ace with 10 cards. [duplicate]
This question already has an answer here:
From a deck of 52 cards, extract 10. In how many combinations do you get at least one ace?
2 answers
We have deck of cards (52 cards). 10 random cards will be picked. How many chances are there, that atleast 1 of the random cards is "ace".
At first I tried to calculate how many different possibilities are there to pull 10 random cards without any extra conditions.
According to formula nCk= n! / k!(n-k)!:
I got 52! / 10! * 42!
Next I tried to calculate how many possibilities are to pull 10 cards from deck, that does not contain any "aces"
I got 48! / 10! * 38!
Then I could subtract second answer from first and I would be left over with answer. Since the numbers are too huge to try on calculator, I wondered if my solutions works.
combinatorics
asked Nov 25 at 19:27
GWL
63
marked as duplicate by N. F. Taussig combinatorics
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Nov 27 at 23:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
From a deck of 52 cards, extract 10. In how many combinations do you get at least one ace?
2 answers
We have deck of cards (52 cards). 10 random cards will be picked. How many chances are there, that atleast 1 of the random cards is "ace".
At first I tried to calculate how many different possibilities are there to pull 10 random cards without any extra conditions.
According to formula nCk= n! / k!(n-k)!:
I got 52! / 10! * 42!
Next I tried to calculate how many possibilities are to pull 10 cards from deck, that does not contain any "aces"
I got 48! / 10! * 38!
Then I could subtract second answer from first and I would be left over with answer. Since the numbers are too huge to try on calculator, I wondered if my solutions works.
combinatorics
asked Nov 25 at 19:27
GWL
63
marked as duplicate by N. F. Taussig combinatorics
Users with the combinatorics badge can single-handedly close combinatorics questions as duplicates and reopen them as needed.
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Nov 27 at 23:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
The logic looks good. Note: since you are interested in the ratio, there is a great deal of cancellation. There's no need to compute either binomial coefficient explicitly.
– lulu
Nov 25 at 19:28
As a different approach: the probability that the first is not an ace is $frac {48}{52}$. Conditioned on that, the probability that the second is also not an ace is $frac {47}{51}$. And so on.
– lulu
Nov 25 at 19:30
@lulu wait. As I understand you are answering, what is probability to get an ace. I need to know how many different combinations are there, that contain ace.
– GWL
Nov 25 at 19:39
I understood "how many chances" to mean "what's the probability". If you just meant the number of combinations, then you can use the binomial symbol approach or you can multiply my expression by the number of unrestricted combinations.
– lulu
Nov 25 at 19:41
add a comment |
This question already has an answer here:
From a deck of 52 cards, extract 10. In how many combinations do you get at least one ace?
2 answers
We have deck of cards (52 cards). 10 random cards will be picked. How many chances are there, that atleast 1 of the random cards is "ace".
At first I tried to calculate how many different possibilities are there to pull 10 random cards without any extra conditions.
According to formula nCk= n! / k!(n-k)!:
I got 52! / 10! * 42!
Next I tried to calculate how many possibilities are to pull 10 cards from deck, that does not contain any "aces"
I got 48! / 10! * 38!
Then I could subtract second answer from first and I would be left over with answer. Since the numbers are too huge to try on calculator, I wondered if my solutions works.
combinatorics
asked Nov 25 at 19:27
GWL
63
This question already has an answer here:
From a deck of 52 cards, extract 10. In how many combinations do you get at least one ace?
2 answers
We have deck of cards (52 cards). 10 random cards will be picked. How many chances are there, that atleast 1 of the random cards is "ace".
At first I tried to calculate how many different possibilities are there to pull 10 random cards without any extra conditions.
According to formula nCk= n! / k!(n-k)!:
I got 52! / 10! * 42!
Next I tried to calculate how many possibilities are to pull 10 cards from deck, that does not contain any "aces"
I got 48! / 10! * 38!
Then I could subtract second answer from first and I would be left over with answer. Since the numbers are too huge to try on calculator, I wondered if my solutions works.
This question already has an answer here:
From a deck of 52 cards, extract 10. In how many combinations do you get at least one ace?
2 answers
combinatorics
combinatorics
asked Nov 25 at 19:27
GWL
63
asked Nov 25 at 19:27
GWL
63
edited Nov 25 at 19:41
asked Nov 25 at 19:27
GWL
63
asked Nov 25 at 19:27
GWL
63
asked Nov 25 at 19:27
GWL
63
63
marked as duplicate by N. F. Taussig combinatorics
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Nov 27 at 23:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by N. F. Taussig combinatorics
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Nov 27 at 23:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
The logic looks good. Note: since you are interested in the ratio, there is a great deal of cancellation. There's no need to compute either binomial coefficient explicitly.
– lulu
Nov 25 at 19:28
As a different approach: the probability that the first is not an ace is $frac {48}{52}$. Conditioned on that, the probability that the second is also not an ace is $frac {47}{51}$. And so on.
– lulu
Nov 25 at 19:30
@lulu wait. As I understand you are answering, what is probability to get an ace. I need to know how many different combinations are there, that contain ace.
– GWL
Nov 25 at 19:39
I understood "how many chances" to mean "what's the probability". If you just meant the number of combinations, then you can use the binomial symbol approach or you can multiply my expression by the number of unrestricted combinations.
– lulu
Nov 25 at 19:41
add a comment |
2
The logic looks good. Note: since you are interested in the ratio, there is a great deal of cancellation. There's no need to compute either binomial coefficient explicitly.
– lulu
Nov 25 at 19:28
As a different approach: the probability that the first is not an ace is $frac {48}{52}$. Conditioned on that, the probability that the second is also not an ace is $frac {47}{51}$. And so on.
– lulu
Nov 25 at 19:30
@lulu wait. As I understand you are answering, what is probability to get an ace. I need to know how many different combinations are there, that contain ace.
– GWL
Nov 25 at 19:39
I understood "how many chances" to mean "what's the probability". If you just meant the number of combinations, then you can use the binomial symbol approach or you can multiply my expression by the number of unrestricted combinations.
– lulu
Nov 25 at 19:41
2
2
The logic looks good. Note: since you are interested in the ratio, there is a great deal of cancellation. There's no need to compute either binomial coefficient explicitly.
– lulu
Nov 25 at 19:28
The logic looks good. Note: since you are interested in the ratio, there is a great deal of cancellation. There's no need to compute either binomial coefficient explicitly.
– lulu
Nov 25 at 19:28
As a different approach: the probability that the first is not an ace is $frac {48}{52}$. Conditioned on that, the probability that the second is also not an ace is $frac {47}{51}$. And so on.
– lulu
Nov 25 at 19:30
As a different approach: the probability that the first is not an ace is $frac {48}{52}$. Conditioned on that, the probability that the second is also not an ace is $frac {47}{51}$. And so on.
– lulu
Nov 25 at 19:30
@lulu wait. As I understand you are answering, what is probability to get an ace. I need to know how many different combinations are there, that contain ace.
– GWL
Nov 25 at 19:39
@lulu wait. As I understand you are answering, what is probability to get an ace. I need to know how many different combinations are there, that contain ace.
– GWL
Nov 25 at 19:39
I understood "how many chances" to mean "what's the probability". If you just meant the number of combinations, then you can use the binomial symbol approach or you can multiply my expression by the number of unrestricted combinations.
– lulu
Nov 25 at 19:41
I understood "how many chances" to mean "what's the probability". If you just meant the number of combinations, then you can use the binomial symbol approach or you can multiply my expression by the number of unrestricted combinations.
– lulu
Nov 25 at 19:41
add a comment |
1 Answer
1
active
oldest
votes
It's okay to give the answer as:
${52choose 10} - {48choose 10}$
That is an acceptable answer but if you want to caluculate it it is.
$frac {52!}{42!10!} - frac {48!}{38!10!} =$
$frac {48!}{38!10!}(frac {52*51*50*49}{42*41*40*39} - 1)=$
$frac {39*....*48}{10!}(frac {52*51*50*49}{42*41*40*39}-frac {42*41*40*39}{42*41*40*39})=$
$frac {43*44*....*48}{10!}(52*51*50*49 - 42*41*40*39)=$
.... factor terms out.
$frac {(43*22*1*46*47*1)*6*8*9*10}{10!}(13*17*5*7 - 1*41*2*39)2*3*4*5*7=$
$(43*22*46*47)(13*17*5*7 - 41*2*39)$
And that can be plugged into a calculator.
But.... 1) NOBODY CARES!!!! The answer ${52choose 10} - {48choose 10}$ is good enough and
2) If you have a computer it comes with a calculator and the numbers aren't too high
$52! = 8.0658175170943878571660636856404e+67$
$frac {52!}{42!} = 57407703889536000$
$frac {52!}{42!10!} = 15820024220$
$48! = 1.2413915592536072670862289047373e+61$
$frac {48!}{38!10!} = 6540715896$
and $frac {52!}{42!10!} - frac {48}{38!10!} = 9279308324$
.... which.... nobody cares. Unless you are doing probability ... in which case $frac {{52choose 10} - {48 choose 10}}{52choose 10}$ is easy to calculate with cancelation as $1 -frac {48!42!}{52!38!}= 1- frac {39*40*41*42}{49*50*51*52} = 1- frac{41*6}{7*5*17}$
answered Nov 27 at 18:01
fleablood
68k22684
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's okay to give the answer as:
${52choose 10} - {48choose 10}$
That is an acceptable answer but if you want to caluculate it it is.
$frac {52!}{42!10!} - frac {48!}{38!10!} =$
$frac {48!}{38!10!}(frac {52*51*50*49}{42*41*40*39} - 1)=$
$frac {39*....*48}{10!}(frac {52*51*50*49}{42*41*40*39}-frac {42*41*40*39}{42*41*40*39})=$
$frac {43*44*....*48}{10!}(52*51*50*49 - 42*41*40*39)=$
.... factor terms out.
$frac {(43*22*1*46*47*1)*6*8*9*10}{10!}(13*17*5*7 - 1*41*2*39)2*3*4*5*7=$
$(43*22*46*47)(13*17*5*7 - 41*2*39)$
And that can be plugged into a calculator.
But.... 1) NOBODY CARES!!!! The answer ${52choose 10} - {48choose 10}$ is good enough and
2) If you have a computer it comes with a calculator and the numbers aren't too high
$52! = 8.0658175170943878571660636856404e+67$
$frac {52!}{42!} = 57407703889536000$
$frac {52!}{42!10!} = 15820024220$
$48! = 1.2413915592536072670862289047373e+61$
$frac {48!}{38!10!} = 6540715896$
and $frac {52!}{42!10!} - frac {48}{38!10!} = 9279308324$
.... which.... nobody cares. Unless you are doing probability ... in which case $frac {{52choose 10} - {48 choose 10}}{52choose 10}$ is easy to calculate with cancelation as $1 -frac {48!42!}{52!38!}= 1- frac {39*40*41*42}{49*50*51*52} = 1- frac{41*6}{7*5*17}$
answered Nov 27 at 18:01
fleablood
68k22684
add a comment |
It's okay to give the answer as:
${52choose 10} - {48choose 10}$
That is an acceptable answer but if you want to caluculate it it is.
$frac {52!}{42!10!} - frac {48!}{38!10!} =$
$frac {48!}{38!10!}(frac {52*51*50*49}{42*41*40*39} - 1)=$
$frac {39*....*48}{10!}(frac {52*51*50*49}{42*41*40*39}-frac {42*41*40*39}{42*41*40*39})=$
$frac {43*44*....*48}{10!}(52*51*50*49 - 42*41*40*39)=$
.... factor terms out.
$frac {(43*22*1*46*47*1)*6*8*9*10}{10!}(13*17*5*7 - 1*41*2*39)2*3*4*5*7=$
$(43*22*46*47)(13*17*5*7 - 41*2*39)$
And that can be plugged into a calculator.
But.... 1) NOBODY CARES!!!! The answer ${52choose 10} - {48choose 10}$ is good enough and
2) If you have a computer it comes with a calculator and the numbers aren't too high
$52! = 8.0658175170943878571660636856404e+67$
$frac {52!}{42!} = 57407703889536000$
$frac {52!}{42!10!} = 15820024220$
$48! = 1.2413915592536072670862289047373e+61$
$frac {48!}{38!10!} = 6540715896$
and $frac {52!}{42!10!} - frac {48}{38!10!} = 9279308324$
.... which.... nobody cares. Unless you are doing probability ... in which case $frac {{52choose 10} - {48 choose 10}}{52choose 10}$ is easy to calculate with cancelation as $1 -frac {48!42!}{52!38!}= 1- frac {39*40*41*42}{49*50*51*52} = 1- frac{41*6}{7*5*17}$
answered Nov 27 at 18:01
fleablood
68k22684
add a comment |
It's okay to give the answer as:
${52choose 10} - {48choose 10}$
That is an acceptable answer but if you want to caluculate it it is.
$frac {52!}{42!10!} - frac {48!}{38!10!} =$
$frac {48!}{38!10!}(frac {52*51*50*49}{42*41*40*39} - 1)=$
$frac {39*....*48}{10!}(frac {52*51*50*49}{42*41*40*39}-frac {42*41*40*39}{42*41*40*39})=$
$frac {43*44*....*48}{10!}(52*51*50*49 - 42*41*40*39)=$
.... factor terms out.
$frac {(43*22*1*46*47*1)*6*8*9*10}{10!}(13*17*5*7 - 1*41*2*39)2*3*4*5*7=$
$(43*22*46*47)(13*17*5*7 - 41*2*39)$
And that can be plugged into a calculator.
But.... 1) NOBODY CARES!!!! The answer ${52choose 10} - {48choose 10}$ is good enough and
2) If you have a computer it comes with a calculator and the numbers aren't too high
$52! = 8.0658175170943878571660636856404e+67$
$frac {52!}{42!} = 57407703889536000$
$frac {52!}{42!10!} = 15820024220$
$48! = 1.2413915592536072670862289047373e+61$
$frac {48!}{38!10!} = 6540715896$
and $frac {52!}{42!10!} - frac {48}{38!10!} = 9279308324$
.... which.... nobody cares. Unless you are doing probability ... in which case $frac {{52choose 10} - {48 choose 10}}{52choose 10}$ is easy to calculate with cancelation as $1 -frac {48!42!}{52!38!}= 1- frac {39*40*41*42}{49*50*51*52} = 1- frac{41*6}{7*5*17}$
answered Nov 27 at 18:01
fleablood
68k22684
It's okay to give the answer as:
${52choose 10} - {48choose 10}$
That is an acceptable answer but if you want to caluculate it it is.
$frac {52!}{42!10!} - frac {48!}{38!10!} =$
$frac {48!}{38!10!}(frac {52*51*50*49}{42*41*40*39} - 1)=$
$frac {39*....*48}{10!}(frac {52*51*50*49}{42*41*40*39}-frac {42*41*40*39}{42*41*40*39})=$
$frac {43*44*....*48}{10!}(52*51*50*49 - 42*41*40*39)=$
.... factor terms out.
$frac {(43*22*1*46*47*1)*6*8*9*10}{10!}(13*17*5*7 - 1*41*2*39)2*3*4*5*7=$
$(43*22*46*47)(13*17*5*7 - 41*2*39)$
And that can be plugged into a calculator.
But.... 1) NOBODY CARES!!!! The answer ${52choose 10} - {48choose 10}$ is good enough and
2) If you have a computer it comes with a calculator and the numbers aren't too high
$52! = 8.0658175170943878571660636856404e+67$
$frac {52!}{42!} = 57407703889536000$
$frac {52!}{42!10!} = 15820024220$
$48! = 1.2413915592536072670862289047373e+61$
$frac {48!}{38!10!} = 6540715896$
and $frac {52!}{42!10!} - frac {48}{38!10!} = 9279308324$
.... which.... nobody cares. Unless you are doing probability ... in which case $frac {{52choose 10} - {48 choose 10}}{52choose 10}$ is easy to calculate with cancelation as $1 -frac {48!42!}{52!38!}= 1- frac {39*40*41*42}{49*50*51*52} = 1- frac{41*6}{7*5*17}$
answered Nov 27 at 18:01
fleablood
68k22684
answered Nov 27 at 18:01
fleablood
68k22684
answered Nov 27 at 18:01
fleablood
68k22684
answered Nov 27 at 18:01
fleablood
68k22684
68k22684
add a comment |
add a comment |
2
The logic looks good. Note: since you are interested in the ratio, there is a great deal of cancellation. There's no need to compute either binomial coefficient explicitly.
– lulu
Nov 25 at 19:28
As a different approach: the probability that the first is not an ace is $frac {48}{52}$. Conditioned on that, the probability that the second is also not an ace is $frac {47}{51}$. And so on.
– lulu
Nov 25 at 19:30
@lulu wait. As I understand you are answering, what is probability to get an ace. I need to know how many different combinations are there, that contain ace.
– GWL
Nov 25 at 19:39
I understood "how many chances" to mean "what's the probability". If you just meant the number of combinations, then you can use the binomial symbol approach or you can multiply my expression by the number of unrestricted combinations.
– lulu
Nov 25 at 19:41