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How many ways can 200 identical balls be distributed into 40 distinct jars?

Distinct balls into distinct boxes with a minimal number of balls in each boxNumber of ways of choosing identical ballsHow many ways are there to place 7 distinct balls into 3 distinct boxes?Ways to distribute $400$ non distinct balls to $3$ bins such that in all of the bins there are more than $200$ balls or less than $100$ balls$12$ Identical balls can be placed into $3$ identical boxes,Total possible ways to arrange 40 identical balls in 3 different boxesChoosing distinct balls to put into indentical urns.How many ways are there to distribute 26 identical balls into six distinct boxes such that…Counting distributions of distinct balls into distinct boxes without PIEDistributing $x$ identical balls into $p$ distinct boxes, what is the probability that there are at most $n$ balls in any of the boxes?

6

1

$begingroup$

How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?

I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $binom23940$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $binom11920^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is

$$fracbinom23940 - binom11920^22$$

The second solution uses a sum. It comes out to

$$sum_k=101^200 binomk+1920binom219-k20$$

combinatorics

share|cite|improve this question

edited Mar 16 at 13:30

YuiTo Cheng

2,1212837

asked Mar 16 at 8:46

David rossDavid ross

311

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – José Carlos Santos
  Mar 16 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
  $endgroup$
  – Yanior Weg
  Mar 16 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YaniorWeg $C^x_y$ that is I think he means combination.
  $endgroup$
  – Bijayan Ray
  Mar 16 at 8:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is not the number of possible ways to place 200 balls in 40 jars $binom23939$?
  $endgroup$
  – user
  Mar 16 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user Still waking up. Yes, you are correct.
  $endgroup$
  – N. F. Taussig
  Mar 16 at 10:26
  

  
  
  
  

 | 
show 2 more comments

6

1

$begingroup$

How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?

I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $binom23940$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $binom11920^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is

$$fracbinom23940 - binom11920^22$$

The second solution uses a sum. It comes out to

$$sum_k=101^200 binomk+1920binom219-k20$$

combinatorics

share|cite|improve this question

edited Mar 16 at 13:30

YuiTo Cheng

2,1212837

asked Mar 16 at 8:46

David rossDavid ross

311

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – José Carlos Santos
  Mar 16 at 8:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
  $endgroup$
  – Yanior Weg
  Mar 16 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @YaniorWeg $C^x_y$ that is I think he means combination.
  $endgroup$
  – Bijayan Ray
  Mar 16 at 8:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is not the number of possible ways to place 200 balls in 40 jars $binom23939$?
  $endgroup$
  – user
  Mar 16 at 10:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user Still waking up. Yes, you are correct.
  $endgroup$
  – N. F. Taussig
  Mar 16 at 10:26
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

How many ways can $200$ identical balls be distributed into $40$ distinct jars so that number Balls in the first $20$ jars is greater than the number of balls in the last $20$ jars?

I came up with two different solutions to the problem but the answers come out differently. The first solution uses symmetry:
There are $binom23940$ possible ways to place $200$ balls in $40$ jars. We then subtract all possibilities where the number of balls in the first $20$ jars is equal to the number of balls in the last $20$ jars, which is equal to $binom11920^2$. We then exploit symmetry (the number of possibilities that there are more balls in the first $20$ jars is equal to the number of possibilities that there are more balls in the last twenty jars). The final answer is

$$fracbinom23940 - binom11920^22$$

The second solution uses a sum. It comes out to

$$sum_k=101^200 binomk+1920binom219-k20$$

combinatorics

share|cite|improve this question

edited Mar 16 at 13:30

YuiTo Cheng

2,1212837

asked Mar 16 at 8:46

David rossDavid ross

311

$endgroup$

share|cite|improve this question

edited Mar 16 at 13:30

YuiTo Cheng

2,1212837

asked Mar 16 at 8:46

David rossDavid ross

311

share|cite|improve this question

edited Mar 16 at 13:30

YuiTo Cheng

2,1212837

asked Mar 16 at 8:46

David rossDavid ross

311

edited Mar 16 at 13:30

YuiTo Cheng

2,1212837

edited Mar 16 at 13:30

YuiTo Cheng

2,1212837

asked Mar 16 at 8:46

David rossDavid ross

311

asked Mar 16 at 8:46

David rossDavid ross

311

2 Answers
2

active

oldest

votes

3

$begingroup$

An expression for the number found by means of stars and bars is:
$$sum_k=0^99binomk+1919binom200-k+1919$$
We can rewrite this as:$$sum_i+j=238wedge ileq118binomi19binomj19$$under the convention that $binomnm=0$ if $mnotin0,1,dots,n$.

Further we have:$$binom23939=sum_i+j=238binomi19binomj19=$$$$sum_i+j=238wedge ileq118binomi19binomj19+sum_i+j=238wedge jleq118binomi19binomj19+binom11919^2$$where the first equality can be recognized as the hockey-stick equality.

This with:$$sum_i+j=238wedge ileq118binomi19binomj19=sum_i+j=238wedge jleq118binomi19binomj19$$
so that:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$

share|cite|improve this answer

answered Mar 16 at 10:21

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
  $endgroup$
  – Eric Duminil
  Mar 16 at 11:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the help
  $endgroup$
  – David ross
  Mar 17 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are very welcome.
  $endgroup$
  – drhab
  Mar 17 at 11:48
  

  
  
  
  

add a comment | 

3

$begingroup$

You made an error in applying "stars and bars". You should replace $binom23940$ with $binom23939$, and $binom11920$ with $binom11919$ and so on.

With correct expressions you obtain:
$$
sum_k=101^200 binomk+1919binom219-k19=fracbinom23939 - binom11919^22.
$$

No contradiction appears.

share|cite|improve this answer

answered Mar 16 at 10:26

useruser

5,82311031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I must have just gotten confused with the n and k notation
  $endgroup$
  – David ross
  Mar 17 at 9:28
  

  
  
  
  

add a comment | 

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3

$begingroup$

An expression for the number found by means of stars and bars is:
$$sum_k=0^99binomk+1919binom200-k+1919$$
We can rewrite this as:$$sum_i+j=238wedge ileq118binomi19binomj19$$under the convention that $binomnm=0$ if $mnotin0,1,dots,n$.

Further we have:$$binom23939=sum_i+j=238binomi19binomj19=$$$$sum_i+j=238wedge ileq118binomi19binomj19+sum_i+j=238wedge jleq118binomi19binomj19+binom11919^2$$where the first equality can be recognized as the hockey-stick equality.

This with:$$sum_i+j=238wedge ileq118binomi19binomj19=sum_i+j=238wedge jleq118binomi19binomj19$$
so that:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$

share|cite|improve this answer

answered Mar 16 at 10:21

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
  $endgroup$
  – Eric Duminil
  Mar 16 at 11:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the help
  $endgroup$
  – David ross
  Mar 17 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are very welcome.
  $endgroup$
  – drhab
  Mar 17 at 11:48
  

  
  
  
  

add a comment | 

3

$begingroup$

An expression for the number found by means of stars and bars is:
$$sum_k=0^99binomk+1919binom200-k+1919$$
We can rewrite this as:$$sum_i+j=238wedge ileq118binomi19binomj19$$under the convention that $binomnm=0$ if $mnotin0,1,dots,n$.

Further we have:$$binom23939=sum_i+j=238binomi19binomj19=$$$$sum_i+j=238wedge ileq118binomi19binomj19+sum_i+j=238wedge jleq118binomi19binomj19+binom11919^2$$where the first equality can be recognized as the hockey-stick equality.

This with:$$sum_i+j=238wedge ileq118binomi19binomj19=sum_i+j=238wedge jleq118binomi19binomj19$$
so that:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$

share|cite|improve this answer

answered Mar 16 at 10:21

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
  $endgroup$
  – Eric Duminil
  Mar 16 at 11:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the help
  $endgroup$
  – David ross
  Mar 17 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are very welcome.
  $endgroup$
  – drhab
  Mar 17 at 11:48
  

  
  
  
  

add a comment | 

$begingroup$

An expression for the number found by means of stars and bars is:
$$sum_k=0^99binomk+1919binom200-k+1919$$
We can rewrite this as:$$sum_i+j=238wedge ileq118binomi19binomj19$$under the convention that $binomnm=0$ if $mnotin0,1,dots,n$.

Further we have:$$binom23939=sum_i+j=238binomi19binomj19=$$$$sum_i+j=238wedge ileq118binomi19binomj19+sum_i+j=238wedge jleq118binomi19binomj19+binom11919^2$$where the first equality can be recognized as the hockey-stick equality.

This with:$$sum_i+j=238wedge ileq118binomi19binomj19=sum_i+j=238wedge jleq118binomi19binomj19$$
so that:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$

share|cite|improve this answer

answered Mar 16 at 10:21

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

answered Mar 16 at 10:21

drhabdrhab

103k545136

answered Mar 16 at 10:21

drhabdrhab

103k545136

answered Mar 16 at 10:21

drhabdrhab

103k545136

3

$begingroup$

You made an error in applying "stars and bars". You should replace $binom23940$ with $binom23939$, and $binom11920$ with $binom11919$ and so on.

With correct expressions you obtain:
$$
sum_k=101^200 binomk+1919binom219-k19=fracbinom23939 - binom11919^22.
$$

No contradiction appears.

share|cite|improve this answer

answered Mar 16 at 10:26

useruser

5,82311031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I must have just gotten confused with the n and k notation
  $endgroup$
  – David ross
  Mar 17 at 9:28
  

  
  
  
  

add a comment | 

3

$begingroup$

You made an error in applying "stars and bars". You should replace $binom23940$ with $binom23939$, and $binom11920$ with $binom11919$ and so on.

With correct expressions you obtain:
$$
sum_k=101^200 binomk+1919binom219-k19=fracbinom23939 - binom11919^22.
$$

No contradiction appears.

share|cite|improve this answer

answered Mar 16 at 10:26

useruser

5,82311031

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I must have just gotten confused with the n and k notation
  $endgroup$
  – David ross
  Mar 17 at 9:28
  

  
  
  
  

add a comment | 

$begingroup$

You made an error in applying "stars and bars". You should replace $binom23940$ with $binom23939$, and $binom11920$ with $binom11919$ and so on.

With correct expressions you obtain:
$$
sum_k=101^200 binomk+1919binom219-k19=fracbinom23939 - binom11919^22.
$$

No contradiction appears.

share|cite|improve this answer

answered Mar 16 at 10:26

useruser

5,82311031

$endgroup$

share|cite|improve this answer

answered Mar 16 at 10:26

useruser

5,82311031

answered Mar 16 at 10:26

useruser

5,82311031

answered Mar 16 at 10:26

useruser

5,82311031