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How many combinations to put books in $n$ packages - Whats wrong with my solution

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0

$begingroup$

There are $2n$ science fiction books, $3n$ thrillers and $5n$ drama.

We want to fix $n$ packages of books so that in each package there will be 2 sci-fi, 3 thrillers and 5 drama.

How many combinations are to do that?

I thought about organising let's say the sci-fi books in a row $(2n)!$ and take every 2 as one, so putting in order in $n$ packages is $n!$.
Same with the other genres so that the solution is $$(2n)!n!(3n)!n!(5n)!n!$$
But not sure if it's the right solution.

What do you think?
How can I know if this is a correct approach?

---

Correct me if I'm wrong, I think my mistake is that I don't need the $(n!)^3)$ I wrote cuz it counts inside the $(2n)!(3n)!(5n)!$,
But I do need to divide with the inner order of each group of 2's, 3's and 5's means divide by $(2!)^n(3!)^(5!)^n$

combinatorics

share|cite|improve this question

edited Mar 29 at 12:46

Evy

asked Mar 29 at 10:25

EvyEvy

244

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In your edit you are correct if the packages are ordered, but I suspect they are not. Then division with $n!$ repairs that.
  $endgroup$
  – drhab
  Mar 29 at 13:30
  

  
  
  
  

add a comment | 

0

$begingroup$

There are $2n$ science fiction books, $3n$ thrillers and $5n$ drama.

We want to fix $n$ packages of books so that in each package there will be 2 sci-fi, 3 thrillers and 5 drama.

How many combinations are to do that?

I thought about organising let's say the sci-fi books in a row $(2n)!$ and take every 2 as one, so putting in order in $n$ packages is $n!$.
Same with the other genres so that the solution is $$(2n)!n!(3n)!n!(5n)!n!$$
But not sure if it's the right solution.

What do you think?
How can I know if this is a correct approach?

---

Correct me if I'm wrong, I think my mistake is that I don't need the $(n!)^3)$ I wrote cuz it counts inside the $(2n)!(3n)!(5n)!$,
But I do need to divide with the inner order of each group of 2's, 3's and 5's means divide by $(2!)^n(3!)^(5!)^n$

combinatorics

share|cite|improve this question

edited Mar 29 at 12:46

Evy

asked Mar 29 at 10:25

EvyEvy

244

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In your edit you are correct if the packages are ordered, but I suspect they are not. Then division with $n!$ repairs that.
  $endgroup$
  – drhab
  Mar 29 at 13:30
  

  
  
  
  

add a comment | 

$begingroup$

There are $2n$ science fiction books, $3n$ thrillers and $5n$ drama.

We want to fix $n$ packages of books so that in each package there will be 2 sci-fi, 3 thrillers and 5 drama.

How many combinations are to do that?

I thought about organising let's say the sci-fi books in a row $(2n)!$ and take every 2 as one, so putting in order in $n$ packages is $n!$.
Same with the other genres so that the solution is $$(2n)!n!(3n)!n!(5n)!n!$$
But not sure if it's the right solution.

What do you think?
How can I know if this is a correct approach?

---

Correct me if I'm wrong, I think my mistake is that I don't need the $(n!)^3)$ I wrote cuz it counts inside the $(2n)!(3n)!(5n)!$,
But I do need to divide with the inner order of each group of 2's, 3's and 5's means divide by $(2!)^n(3!)^(5!)^n$

combinatorics

share|cite|improve this question

edited Mar 29 at 12:46

Evy

asked Mar 29 at 10:25

EvyEvy

244

$endgroup$

share|cite|improve this question

edited Mar 29 at 12:46

Evy

asked Mar 29 at 10:25

EvyEvy

244

share|cite|improve this question

edited Mar 29 at 12:46

Evy

asked Mar 29 at 10:25

EvyEvy

244

asked Mar 29 at 10:25

EvyEvy

244

asked Mar 29 at 10:25

EvyEvy

244

1 Answer
1

active

oldest

votes

4

$begingroup$

Let us first label the packings by $1,2,dots n$.

We start with selecting $2$ sci-fi's, $3$ thrillers and $5$ drama's meant for the first package.

For that there are $binom2n2binom3n3binom5n5$ possibilities.

Then for the second package there are $binom2n-22binom3n-33binom5n-55$ possibilities et cetera, so - if the order of the packages counts - then there are: $$binom2n2binom3n3binom5n5timesbinom2n-22binom3n-33binom5n-55timescdotstimesbinom22binom33binom55=$$$$frac(2n)!(3n)!(5n)!(2!)^n(3!)^n(5!)^n=frac(2n)!(3n)!(5n)!1440^n$$possibilities.

If the order of the packages does not count (and I suspect that is the case here) then every possibility has been counted $n!$ times above, and repairing this we find:$$frac(2n)!(3n)!(5n)!n!1440^n$$possibilities.

share|cite|improve this answer

edited Mar 29 at 13:10

answered Mar 29 at 10:49

drhabdrhab

104k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, can you please explain what's wrong with my solution
  $endgroup$
  – Evy
  Mar 29 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, Should'nt it be $(2!)^n(3!)^n(5!)^n$
  $endgroup$
  – Evy
  Mar 29 at 12:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is what it should be (I repaired).
  $endgroup$
  – drhab
  Mar 29 at 13:15
  
  

  
  
  
  

add a comment | 

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4

$begingroup$

Let us first label the packings by $1,2,dots n$.

We start with selecting $2$ sci-fi's, $3$ thrillers and $5$ drama's meant for the first package.

For that there are $binom2n2binom3n3binom5n5$ possibilities.

Then for the second package there are $binom2n-22binom3n-33binom5n-55$ possibilities et cetera, so - if the order of the packages counts - then there are: $$binom2n2binom3n3binom5n5timesbinom2n-22binom3n-33binom5n-55timescdotstimesbinom22binom33binom55=$$$$frac(2n)!(3n)!(5n)!(2!)^n(3!)^n(5!)^n=frac(2n)!(3n)!(5n)!1440^n$$possibilities.

If the order of the packages does not count (and I suspect that is the case here) then every possibility has been counted $n!$ times above, and repairing this we find:$$frac(2n)!(3n)!(5n)!n!1440^n$$possibilities.

share|cite|improve this answer

edited Mar 29 at 13:10

answered Mar 29 at 10:49

drhabdrhab

104k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, can you please explain what's wrong with my solution
  $endgroup$
  – Evy
  Mar 29 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, Should'nt it be $(2!)^n(3!)^n(5!)^n$
  $endgroup$
  – Evy
  Mar 29 at 12:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is what it should be (I repaired).
  $endgroup$
  – drhab
  Mar 29 at 13:15
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Let us first label the packings by $1,2,dots n$.

We start with selecting $2$ sci-fi's, $3$ thrillers and $5$ drama's meant for the first package.

For that there are $binom2n2binom3n3binom5n5$ possibilities.

Then for the second package there are $binom2n-22binom3n-33binom5n-55$ possibilities et cetera, so - if the order of the packages counts - then there are: $$binom2n2binom3n3binom5n5timesbinom2n-22binom3n-33binom5n-55timescdotstimesbinom22binom33binom55=$$$$frac(2n)!(3n)!(5n)!(2!)^n(3!)^n(5!)^n=frac(2n)!(3n)!(5n)!1440^n$$possibilities.

If the order of the packages does not count (and I suspect that is the case here) then every possibility has been counted $n!$ times above, and repairing this we find:$$frac(2n)!(3n)!(5n)!n!1440^n$$possibilities.

share|cite|improve this answer

edited Mar 29 at 13:10

answered Mar 29 at 10:49

drhabdrhab

104k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, can you please explain what's wrong with my solution
  $endgroup$
  – Evy
  Mar 29 at 11:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, Should'nt it be $(2!)^n(3!)^n(5!)^n$
  $endgroup$
  – Evy
  Mar 29 at 12:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is what it should be (I repaired).
  $endgroup$
  – drhab
  Mar 29 at 13:15
  
  

  
  
  
  

add a comment | 

$begingroup$

Let us first label the packings by $1,2,dots n$.

We start with selecting $2$ sci-fi's, $3$ thrillers and $5$ drama's meant for the first package.

For that there are $binom2n2binom3n3binom5n5$ possibilities.

Then for the second package there are $binom2n-22binom3n-33binom5n-55$ possibilities et cetera, so - if the order of the packages counts - then there are: $$binom2n2binom3n3binom5n5timesbinom2n-22binom3n-33binom5n-55timescdotstimesbinom22binom33binom55=$$$$frac(2n)!(3n)!(5n)!(2!)^n(3!)^n(5!)^n=frac(2n)!(3n)!(5n)!1440^n$$possibilities.

If the order of the packages does not count (and I suspect that is the case here) then every possibility has been counted $n!$ times above, and repairing this we find:$$frac(2n)!(3n)!(5n)!n!1440^n$$possibilities.

share|cite|improve this answer

edited Mar 29 at 13:10

answered Mar 29 at 10:49

drhabdrhab

104k545136

$endgroup$

share|cite|improve this answer

edited Mar 29 at 13:10

answered Mar 29 at 10:49

drhabdrhab

104k545136

answered Mar 29 at 10:49

drhabdrhab

104k545136

answered Mar 29 at 10:49

drhabdrhab

104k545136