How many ways can 200 identical balls be distributed into 40 distinct jars?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
asked Mar 16 at 8:46
David rossDavid ross
412
$endgroup$
|
show 1 more comment
$begingroup$
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
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
asked Mar 16 at 8:46
David rossDavid ross
412
$endgroup$
$begingroup$
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
$begingroup$
why "identical" balls and "distinct" jars?
$endgroup$
– mmw
Mar 16 at 10:48
|
show 1 more comment
$begingroup$
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
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
asked Mar 16 at 8:46
David rossDavid ross
412
$endgroup$
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
combinatorics
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
asked Mar 16 at 8:46
David rossDavid ross
412
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
asked Mar 16 at 8:46
David rossDavid ross
412
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
edited Mar 16 at 13:30
YuiTo Cheng
2,40641037
2,40641037
asked Mar 16 at 8:46
David rossDavid ross
412
asked Mar 16 at 8:46
David rossDavid ross
412
asked Mar 16 at 8:46
David rossDavid ross
412
412
$begingroup$
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
$begingroup$
why "identical" balls and "distinct" jars?
$endgroup$
– mmw
Mar 16 at 10:48
|
show 1 more comment
$begingroup$
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
$begingroup$
why "identical" balls and "distinct" jars?
$endgroup$
– mmw
Mar 16 at 10:48
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
Mar 16 at 10:26
$begingroup$
why "identical" balls and "distinct" jars?
$endgroup$
– mmw
Mar 16 at 10:48
$begingroup$
why "identical" balls and "distinct" jars?
$endgroup$
– mmw
Mar 16 at 10:48
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$
answered Mar 16 at 10:21
drhabdrhab
104k545136
$endgroup$
$begingroup$
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 |
$begingroup$
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.
answered Mar 16 at 10:26
useruser
6,57011031
$endgroup$
$begingroup$
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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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
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$begingroup$
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:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$
answered Mar 16 at 10:21
drhabdrhab
104k545136
$endgroup$
$begingroup$
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:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$
answered Mar 16 at 10:21
drhabdrhab
104k545136
$endgroup$
$begingroup$
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:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$
answered Mar 16 at 10:21
drhabdrhab
104k545136
$endgroup$
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:$$sum_i+j=238wedge ileq118binomi19binomj19=frac12left[binom23939-binom11919^2right]$$
answered Mar 16 at 10:21
drhabdrhab
104k545136
answered Mar 16 at 10:21
drhabdrhab
104k545136
answered Mar 16 at 10:21
drhabdrhab
104k545136
answered Mar 16 at 10:21
drhabdrhab
104k545136
104k545136
$begingroup$
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 |
$begingroup$
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
$begingroup$
Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
$endgroup$
– Eric Duminil
Mar 16 at 11:23
$begingroup$
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
$begingroup$
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
$begingroup$
You are very welcome.
$endgroup$
– drhab
Mar 17 at 11:48
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.
answered Mar 16 at 10:26
useruser
6,57011031
$endgroup$
$begingroup$
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.
answered Mar 16 at 10:26
useruser
6,57011031
$endgroup$
$begingroup$
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.
answered Mar 16 at 10:26
useruser
6,57011031
$endgroup$
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.
answered Mar 16 at 10:26
useruser
6,57011031
answered Mar 16 at 10:26
useruser
6,57011031
answered Mar 16 at 10:26
useruser
6,57011031
answered Mar 16 at 10:26
useruser
6,57011031
6,57011031
$begingroup$
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 |
$begingroup$
Thank you, I must have just gotten confused with the n and k notation
$endgroup$
– David ross
Mar 17 at 9:28
$begingroup$
Thank you, I must have just gotten confused with the n and k notation
$endgroup$
– David ross
Mar 17 at 9:28
$begingroup$
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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$begingroup$
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
$begingroup$
why "identical" balls and "distinct" jars?
$endgroup$
– mmw
Mar 16 at 10:48