Combinatorics problem (deck of cards)












4












$begingroup$


I'd be extremely thankful if someone could help me with this.



A company makes decks of $50$ cards. There are $40$ "regular" cards, that always come in a certain order, and $10$ "special" cards that don't have a particular order and also they are interspersed between the deck. (Note the $40$ regular cards don't always come one after the other, they just always follow the same order)



In how many different combinations could the deck come?



My thinking (which apparently is wrong) was to say well, I'll choose $10$ out of the $50$ possible places and multiply that by $10!$ which is the number of ways I can shuffle the special cards. So
$$
binom{50}{10} cdot 10! .
$$



And that would be it since the regular cards will always be positioned in a certain order.
But the answer key says




$$binom{n+k-1}{n} = binom{49}{10}cdot 10! = frac{49!}{39!}.$$




Any clue why instead of picking $10$ out of $50$ it picks $10$ out of $49$?



Thanks.






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would say that 50 doesn't make sense but 51 since you want to insert cards in between and not "in" one card. So you have 49 places in between and 51 if you count the borders.
    $endgroup$
    – sehigle
    Dec 20 '18 at 9:31






  • 5




    $begingroup$
    I think the answer key is wrong and you are right. As a sanity check apply on it on cases like: there are $0$ regular cards on $10$ special cards, or there is $1$ regular card and $10$ special cards.
    $endgroup$
    – drhab
    Dec 20 '18 at 9:44






  • 1




    $begingroup$
    Did you cite the original problem statement word for word?
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:49










  • $begingroup$
    @Christoph Not really because the problem is in spanish, so I did my best to translate it. I guess maybe there's an error in my understanding of the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 9:52






  • 1




    $begingroup$
    Also note that for no choices of $n$ and $k$ we have $binom{n+k-1}{n} = binom{49}{10}cdot 10!$, so something about the given answer is wrong anyway.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:21


















4












$begingroup$


I'd be extremely thankful if someone could help me with this.



A company makes decks of $50$ cards. There are $40$ "regular" cards, that always come in a certain order, and $10$ "special" cards that don't have a particular order and also they are interspersed between the deck. (Note the $40$ regular cards don't always come one after the other, they just always follow the same order)



In how many different combinations could the deck come?



My thinking (which apparently is wrong) was to say well, I'll choose $10$ out of the $50$ possible places and multiply that by $10!$ which is the number of ways I can shuffle the special cards. So
$$
binom{50}{10} cdot 10! .
$$



And that would be it since the regular cards will always be positioned in a certain order.
But the answer key says




$$binom{n+k-1}{n} = binom{49}{10}cdot 10! = frac{49!}{39!}.$$




Any clue why instead of picking $10$ out of $50$ it picks $10$ out of $49$?



Thanks.






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    I would say that 50 doesn't make sense but 51 since you want to insert cards in between and not "in" one card. So you have 49 places in between and 51 if you count the borders.
    $endgroup$
    – sehigle
    Dec 20 '18 at 9:31






  • 5




    $begingroup$
    I think the answer key is wrong and you are right. As a sanity check apply on it on cases like: there are $0$ regular cards on $10$ special cards, or there is $1$ regular card and $10$ special cards.
    $endgroup$
    – drhab
    Dec 20 '18 at 9:44






  • 1




    $begingroup$
    Did you cite the original problem statement word for word?
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:49










  • $begingroup$
    @Christoph Not really because the problem is in spanish, so I did my best to translate it. I guess maybe there's an error in my understanding of the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 9:52






  • 1




    $begingroup$
    Also note that for no choices of $n$ and $k$ we have $binom{n+k-1}{n} = binom{49}{10}cdot 10!$, so something about the given answer is wrong anyway.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:21
















4












4








4





$begingroup$


I'd be extremely thankful if someone could help me with this.



A company makes decks of $50$ cards. There are $40$ "regular" cards, that always come in a certain order, and $10$ "special" cards that don't have a particular order and also they are interspersed between the deck. (Note the $40$ regular cards don't always come one after the other, they just always follow the same order)



In how many different combinations could the deck come?



My thinking (which apparently is wrong) was to say well, I'll choose $10$ out of the $50$ possible places and multiply that by $10!$ which is the number of ways I can shuffle the special cards. So
$$
binom{50}{10} cdot 10! .
$$



And that would be it since the regular cards will always be positioned in a certain order.
But the answer key says




$$binom{n+k-1}{n} = binom{49}{10}cdot 10! = frac{49!}{39!}.$$




Any clue why instead of picking $10$ out of $50$ it picks $10$ out of $49$?



Thanks.






combinatorics







share|cite|improve this question











$endgroup$




I'd be extremely thankful if someone could help me with this.



A company makes decks of $50$ cards. There are $40$ "regular" cards, that always come in a certain order, and $10$ "special" cards that don't have a particular order and also they are interspersed between the deck. (Note the $40$ regular cards don't always come one after the other, they just always follow the same order)



In how many different combinations could the deck come?



My thinking (which apparently is wrong) was to say well, I'll choose $10$ out of the $50$ possible places and multiply that by $10!$ which is the number of ways I can shuffle the special cards. So
$$
binom{50}{10} cdot 10! .
$$



And that would be it since the regular cards will always be positioned in a certain order.
But the answer key says




$$binom{n+k-1}{n} = binom{49}{10}cdot 10! = frac{49!}{39!}.$$




Any clue why instead of picking $10$ out of $50$ it picks $10$ out of $49$?



Thanks.






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 10:06









Prakhar Nagpal

747318




747318










asked Dec 20 '18 at 9:24









DirewolfoxDirewolfox

406




406












  • $begingroup$
    I would say that 50 doesn't make sense but 51 since you want to insert cards in between and not "in" one card. So you have 49 places in between and 51 if you count the borders.
    $endgroup$
    – sehigle
    Dec 20 '18 at 9:31






  • 5




    $begingroup$
    I think the answer key is wrong and you are right. As a sanity check apply on it on cases like: there are $0$ regular cards on $10$ special cards, or there is $1$ regular card and $10$ special cards.
    $endgroup$
    – drhab
    Dec 20 '18 at 9:44






  • 1




    $begingroup$
    Did you cite the original problem statement word for word?
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:49










  • $begingroup$
    @Christoph Not really because the problem is in spanish, so I did my best to translate it. I guess maybe there's an error in my understanding of the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 9:52






  • 1




    $begingroup$
    Also note that for no choices of $n$ and $k$ we have $binom{n+k-1}{n} = binom{49}{10}cdot 10!$, so something about the given answer is wrong anyway.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:21




















  • $begingroup$
    I would say that 50 doesn't make sense but 51 since you want to insert cards in between and not "in" one card. So you have 49 places in between and 51 if you count the borders.
    $endgroup$
    – sehigle
    Dec 20 '18 at 9:31






  • 5




    $begingroup$
    I think the answer key is wrong and you are right. As a sanity check apply on it on cases like: there are $0$ regular cards on $10$ special cards, or there is $1$ regular card and $10$ special cards.
    $endgroup$
    – drhab
    Dec 20 '18 at 9:44






  • 1




    $begingroup$
    Did you cite the original problem statement word for word?
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:49










  • $begingroup$
    @Christoph Not really because the problem is in spanish, so I did my best to translate it. I guess maybe there's an error in my understanding of the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 9:52






  • 1




    $begingroup$
    Also note that for no choices of $n$ and $k$ we have $binom{n+k-1}{n} = binom{49}{10}cdot 10!$, so something about the given answer is wrong anyway.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:21


















$begingroup$
I would say that 50 doesn't make sense but 51 since you want to insert cards in between and not "in" one card. So you have 49 places in between and 51 if you count the borders.
$endgroup$
– sehigle
Dec 20 '18 at 9:31




$begingroup$
I would say that 50 doesn't make sense but 51 since you want to insert cards in between and not "in" one card. So you have 49 places in between and 51 if you count the borders.
$endgroup$
– sehigle
Dec 20 '18 at 9:31




5




5




$begingroup$
I think the answer key is wrong and you are right. As a sanity check apply on it on cases like: there are $0$ regular cards on $10$ special cards, or there is $1$ regular card and $10$ special cards.
$endgroup$
– drhab
Dec 20 '18 at 9:44




$begingroup$
I think the answer key is wrong and you are right. As a sanity check apply on it on cases like: there are $0$ regular cards on $10$ special cards, or there is $1$ regular card and $10$ special cards.
$endgroup$
– drhab
Dec 20 '18 at 9:44




1




1




$begingroup$
Did you cite the original problem statement word for word?
$endgroup$
– Christoph
Dec 20 '18 at 9:49




$begingroup$
Did you cite the original problem statement word for word?
$endgroup$
– Christoph
Dec 20 '18 at 9:49












$begingroup$
@Christoph Not really because the problem is in spanish, so I did my best to translate it. I guess maybe there's an error in my understanding of the problem
$endgroup$
– Direwolfox
Dec 20 '18 at 9:52




$begingroup$
@Christoph Not really because the problem is in spanish, so I did my best to translate it. I guess maybe there's an error in my understanding of the problem
$endgroup$
– Direwolfox
Dec 20 '18 at 9:52




1




1




$begingroup$
Also note that for no choices of $n$ and $k$ we have $binom{n+k-1}{n} = binom{49}{10}cdot 10!$, so something about the given answer is wrong anyway.
$endgroup$
– Christoph
Dec 20 '18 at 10:21






$begingroup$
Also note that for no choices of $n$ and $k$ we have $binom{n+k-1}{n} = binom{49}{10}cdot 10!$, so something about the given answer is wrong anyway.
$endgroup$
– Christoph
Dec 20 '18 at 10:21












1 Answer
1






active

oldest

votes


















2












$begingroup$

The correct answer is-



${41 choose 1} cdot {42 choose 1} cdot {43 choose 1} ... {50 choose 1} = 41cdot 42 cdot 43... 50$



This is so, since once we arrange the $40$ regular cards in the order, there are $41$ places left between them (including the extremes). Now, the first special card can be placed in any of these 41 'slots'.



The next special card now has $42$ slots as it has to be placed between $41$ cards.



Continuing the process yields the answer.



It seems that you are supposed to interpret the problem as only allowing the spaces between the cards and not the extremes. In that case, you would have $39$, slots for the first special card, $40$ for the second and so on.



That gives $39 cdot 40 cdot 41 ... 48$.



Overall the problem is just a bit ambiguous, perhaps due some issues in understanding the problem and translation :)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note that $41cdot42cdotdotscdot50=frac{50!}{40!}=binom{50}{10}10!$ which is OPs answer.
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:58












  • $begingroup$
    @Christoph I think you mean $40 cdot 41 ... 50$. The answer is correct but the reasoing is not. In the end it is simply a matter of interpretation, I think :)
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:00










  • $begingroup$
    You can not interpret "scattered around the deck" as "can't be first or last". Maybe the original statement in spanish is different, but the translation does not allow for this interpretation.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:00










  • $begingroup$
    @Christoph Hey. Sorry, I guess a better translation would be that the special ones are "interspersed" between the other 40. or "intercalated". Maybe that's the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 10:03










  • $begingroup$
    @Christoph See edited answer, I think it should probably be better.
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:03











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The correct answer is-



${41 choose 1} cdot {42 choose 1} cdot {43 choose 1} ... {50 choose 1} = 41cdot 42 cdot 43... 50$



This is so, since once we arrange the $40$ regular cards in the order, there are $41$ places left between them (including the extremes). Now, the first special card can be placed in any of these 41 'slots'.



The next special card now has $42$ slots as it has to be placed between $41$ cards.



Continuing the process yields the answer.



It seems that you are supposed to interpret the problem as only allowing the spaces between the cards and not the extremes. In that case, you would have $39$, slots for the first special card, $40$ for the second and so on.



That gives $39 cdot 40 cdot 41 ... 48$.



Overall the problem is just a bit ambiguous, perhaps due some issues in understanding the problem and translation :)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note that $41cdot42cdotdotscdot50=frac{50!}{40!}=binom{50}{10}10!$ which is OPs answer.
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:58












  • $begingroup$
    @Christoph I think you mean $40 cdot 41 ... 50$. The answer is correct but the reasoing is not. In the end it is simply a matter of interpretation, I think :)
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:00










  • $begingroup$
    You can not interpret "scattered around the deck" as "can't be first or last". Maybe the original statement in spanish is different, but the translation does not allow for this interpretation.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:00










  • $begingroup$
    @Christoph Hey. Sorry, I guess a better translation would be that the special ones are "interspersed" between the other 40. or "intercalated". Maybe that's the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 10:03










  • $begingroup$
    @Christoph See edited answer, I think it should probably be better.
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:03
















2












$begingroup$

The correct answer is-



${41 choose 1} cdot {42 choose 1} cdot {43 choose 1} ... {50 choose 1} = 41cdot 42 cdot 43... 50$



This is so, since once we arrange the $40$ regular cards in the order, there are $41$ places left between them (including the extremes). Now, the first special card can be placed in any of these 41 'slots'.



The next special card now has $42$ slots as it has to be placed between $41$ cards.



Continuing the process yields the answer.



It seems that you are supposed to interpret the problem as only allowing the spaces between the cards and not the extremes. In that case, you would have $39$, slots for the first special card, $40$ for the second and so on.



That gives $39 cdot 40 cdot 41 ... 48$.



Overall the problem is just a bit ambiguous, perhaps due some issues in understanding the problem and translation :)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note that $41cdot42cdotdotscdot50=frac{50!}{40!}=binom{50}{10}10!$ which is OPs answer.
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:58












  • $begingroup$
    @Christoph I think you mean $40 cdot 41 ... 50$. The answer is correct but the reasoing is not. In the end it is simply a matter of interpretation, I think :)
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:00










  • $begingroup$
    You can not interpret "scattered around the deck" as "can't be first or last". Maybe the original statement in spanish is different, but the translation does not allow for this interpretation.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:00










  • $begingroup$
    @Christoph Hey. Sorry, I guess a better translation would be that the special ones are "interspersed" between the other 40. or "intercalated". Maybe that's the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 10:03










  • $begingroup$
    @Christoph See edited answer, I think it should probably be better.
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:03














2












2








2





$begingroup$

The correct answer is-



${41 choose 1} cdot {42 choose 1} cdot {43 choose 1} ... {50 choose 1} = 41cdot 42 cdot 43... 50$



This is so, since once we arrange the $40$ regular cards in the order, there are $41$ places left between them (including the extremes). Now, the first special card can be placed in any of these 41 'slots'.



The next special card now has $42$ slots as it has to be placed between $41$ cards.



Continuing the process yields the answer.



It seems that you are supposed to interpret the problem as only allowing the spaces between the cards and not the extremes. In that case, you would have $39$, slots for the first special card, $40$ for the second and so on.



That gives $39 cdot 40 cdot 41 ... 48$.



Overall the problem is just a bit ambiguous, perhaps due some issues in understanding the problem and translation :)






share|cite|improve this answer











$endgroup$



The correct answer is-



${41 choose 1} cdot {42 choose 1} cdot {43 choose 1} ... {50 choose 1} = 41cdot 42 cdot 43... 50$



This is so, since once we arrange the $40$ regular cards in the order, there are $41$ places left between them (including the extremes). Now, the first special card can be placed in any of these 41 'slots'.



The next special card now has $42$ slots as it has to be placed between $41$ cards.



Continuing the process yields the answer.



It seems that you are supposed to interpret the problem as only allowing the spaces between the cards and not the extremes. In that case, you would have $39$, slots for the first special card, $40$ for the second and so on.



That gives $39 cdot 40 cdot 41 ... 48$.



Overall the problem is just a bit ambiguous, perhaps due some issues in understanding the problem and translation :)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 10:16

























answered Dec 20 '18 at 9:53









Devashish KaushikDevashish Kaushik

570219




570219












  • $begingroup$
    Note that $41cdot42cdotdotscdot50=frac{50!}{40!}=binom{50}{10}10!$ which is OPs answer.
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:58












  • $begingroup$
    @Christoph I think you mean $40 cdot 41 ... 50$. The answer is correct but the reasoing is not. In the end it is simply a matter of interpretation, I think :)
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:00










  • $begingroup$
    You can not interpret "scattered around the deck" as "can't be first or last". Maybe the original statement in spanish is different, but the translation does not allow for this interpretation.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:00










  • $begingroup$
    @Christoph Hey. Sorry, I guess a better translation would be that the special ones are "interspersed" between the other 40. or "intercalated". Maybe that's the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 10:03










  • $begingroup$
    @Christoph See edited answer, I think it should probably be better.
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:03


















  • $begingroup$
    Note that $41cdot42cdotdotscdot50=frac{50!}{40!}=binom{50}{10}10!$ which is OPs answer.
    $endgroup$
    – Christoph
    Dec 20 '18 at 9:58












  • $begingroup$
    @Christoph I think you mean $40 cdot 41 ... 50$. The answer is correct but the reasoing is not. In the end it is simply a matter of interpretation, I think :)
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:00










  • $begingroup$
    You can not interpret "scattered around the deck" as "can't be first or last". Maybe the original statement in spanish is different, but the translation does not allow for this interpretation.
    $endgroup$
    – Christoph
    Dec 20 '18 at 10:00










  • $begingroup$
    @Christoph Hey. Sorry, I guess a better translation would be that the special ones are "interspersed" between the other 40. or "intercalated". Maybe that's the problem
    $endgroup$
    – Direwolfox
    Dec 20 '18 at 10:03










  • $begingroup$
    @Christoph See edited answer, I think it should probably be better.
    $endgroup$
    – Devashish Kaushik
    Dec 20 '18 at 10:03
















$begingroup$
Note that $41cdot42cdotdotscdot50=frac{50!}{40!}=binom{50}{10}10!$ which is OPs answer.
$endgroup$
– Christoph
Dec 20 '18 at 9:58






$begingroup$
Note that $41cdot42cdotdotscdot50=frac{50!}{40!}=binom{50}{10}10!$ which is OPs answer.
$endgroup$
– Christoph
Dec 20 '18 at 9:58














$begingroup$
@Christoph I think you mean $40 cdot 41 ... 50$. The answer is correct but the reasoing is not. In the end it is simply a matter of interpretation, I think :)
$endgroup$
– Devashish Kaushik
Dec 20 '18 at 10:00




$begingroup$
@Christoph I think you mean $40 cdot 41 ... 50$. The answer is correct but the reasoing is not. In the end it is simply a matter of interpretation, I think :)
$endgroup$
– Devashish Kaushik
Dec 20 '18 at 10:00












$begingroup$
You can not interpret "scattered around the deck" as "can't be first or last". Maybe the original statement in spanish is different, but the translation does not allow for this interpretation.
$endgroup$
– Christoph
Dec 20 '18 at 10:00




$begingroup$
You can not interpret "scattered around the deck" as "can't be first or last". Maybe the original statement in spanish is different, but the translation does not allow for this interpretation.
$endgroup$
– Christoph
Dec 20 '18 at 10:00












$begingroup$
@Christoph Hey. Sorry, I guess a better translation would be that the special ones are "interspersed" between the other 40. or "intercalated". Maybe that's the problem
$endgroup$
– Direwolfox
Dec 20 '18 at 10:03




$begingroup$
@Christoph Hey. Sorry, I guess a better translation would be that the special ones are "interspersed" between the other 40. or "intercalated". Maybe that's the problem
$endgroup$
– Direwolfox
Dec 20 '18 at 10:03












$begingroup$
@Christoph See edited answer, I think it should probably be better.
$endgroup$
– Devashish Kaushik
Dec 20 '18 at 10:03




$begingroup$
@Christoph See edited answer, I think it should probably be better.
$endgroup$
– Devashish Kaushik
Dec 20 '18 at 10:03


















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