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












6












$begingroup$



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 $binom{239}{40}$ 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 $binom{119}{20}^2$. 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




$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$




The second solution uses a sum. It comes out to




$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$







combinatorics







share|cite|improve this question









New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







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  • 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
    15 hours ago










  • $begingroup$
    Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
    $endgroup$
    – Yanior Weg
    15 hours ago










  • $begingroup$
    @YaniorWeg $C^x_y$ that is I think he means combination.
    $endgroup$
    – Bijayan Ray
    15 hours ago












  • $begingroup$
    Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
    $endgroup$
    – user
    14 hours ago










  • $begingroup$
    @user Still waking up. Yes, you are correct.
    $endgroup$
    – N. F. Taussig
    14 hours ago
















6












$begingroup$



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 $binom{239}{40}$ 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 $binom{119}{20}^2$. 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




$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$




The second solution uses a sum. It comes out to




$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$







combinatorics







share|cite|improve this question









New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 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
    15 hours ago










  • $begingroup$
    Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
    $endgroup$
    – Yanior Weg
    15 hours ago










  • $begingroup$
    @YaniorWeg $C^x_y$ that is I think he means combination.
    $endgroup$
    – Bijayan Ray
    15 hours ago












  • $begingroup$
    Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
    $endgroup$
    – user
    14 hours ago










  • $begingroup$
    @user Still waking up. Yes, you are correct.
    $endgroup$
    – N. F. Taussig
    14 hours ago














6












6








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 $binom{239}{40}$ 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 $binom{119}{20}^2$. 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




$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$




The second solution uses a sum. It comes out to




$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$







combinatorics







share|cite|improve this question









New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





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 $binom{239}{40}$ 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 $binom{119}{20}^2$. 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




$$frac{binom{239}{40} - binom{119}{20}^2}{2}$$




The second solution uses a sum. It comes out to




$$sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$$







combinatorics




combinatorics






share|cite|improve this question









New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 10 hours ago









YuiTo Cheng

2,0532637




2,0532637






New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 15 hours ago









David rossDavid ross

311




311




New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 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
    15 hours ago










  • $begingroup$
    Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
    $endgroup$
    – Yanior Weg
    15 hours ago










  • $begingroup$
    @YaniorWeg $C^x_y$ that is I think he means combination.
    $endgroup$
    – Bijayan Ray
    15 hours ago












  • $begingroup$
    Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
    $endgroup$
    – user
    14 hours ago










  • $begingroup$
    @user Still waking up. Yes, you are correct.
    $endgroup$
    – N. F. Taussig
    14 hours ago














  • 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
    15 hours ago










  • $begingroup$
    Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
    $endgroup$
    – Yanior Weg
    15 hours ago










  • $begingroup$
    @YaniorWeg $C^x_y$ that is I think he means combination.
    $endgroup$
    – Bijayan Ray
    15 hours ago












  • $begingroup$
    Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
    $endgroup$
    – user
    14 hours ago










  • $begingroup$
    @user Still waking up. Yes, you are correct.
    $endgroup$
    – N. F. Taussig
    14 hours ago








1




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
15 hours ago




$begingroup$
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
15 hours ago












$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago




$begingroup$
Hi, welcome to MSE! Could you explain us, please, what do you mean by "($X$ choose $Y$)"?
$endgroup$
– Yanior Weg
15 hours ago












$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago






$begingroup$
@YaniorWeg $C^x_y$ that is I think he means combination.
$endgroup$
– Bijayan Ray
15 hours ago














$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago




$begingroup$
Is not the number of possible ways to place 200 balls in 40 jars $binom{239}{39}$?
$endgroup$
– user
14 hours ago












$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago




$begingroup$
@user Still waking up. Yes, you are correct.
$endgroup$
– N. F. Taussig
14 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

An expression for the number found by means of stars and bars is:
$$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.



Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.



This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
    $endgroup$
    – Eric Duminil
    13 hours ago



















3












$begingroup$

You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.



With correct expressions you obtain:
$$
sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
$$



No contradiction appears.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    An expression for the number found by means of stars and bars is:
    $$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
    We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.



    Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.



    This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
    so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
      $endgroup$
      – Eric Duminil
      13 hours ago
















    3












    $begingroup$

    An expression for the number found by means of stars and bars is:
    $$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
    We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.



    Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.



    This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
    so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
      $endgroup$
      – Eric Duminil
      13 hours ago














    3












    3








    3





    $begingroup$

    An expression for the number found by means of stars and bars is:
    $$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
    We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.



    Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.



    This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
    so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$






    share|cite|improve this answer









    $endgroup$



    An expression for the number found by means of stars and bars is:
    $$sum_{k=0}^{99}binom{k+19}{19}binom{200-k+19}{19}$$
    We can rewrite this as:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}$$under the convention that $binom{n}{m}=0$ if $mnotin{0,1,dots,n}$.



    Further we have:$$binom{239}{39}=sum_{i+j=238}binom{i}{19}binom{j}{19}=$$$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}+sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}+binom{119}{19}^2$$where the first equality can be recognized as the hockey-stick equality.



    This with:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=sum_{i+j=238wedge jleq118}binom{i}{19}binom{j}{19}$$
    so that:$$sum_{i+j=238wedge ileq118}binom{i}{19}binom{j}{19}=frac{1}{2}left[binom{239}{39}-binom{119}{19}^2right]$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 14 hours ago









    drhabdrhab

    103k545136




    103k545136












    • $begingroup$
      Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
      $endgroup$
      – Eric Duminil
      13 hours ago


















    • $begingroup$
      Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
      $endgroup$
      – Eric Duminil
      13 hours ago
















    $begingroup$
    Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
    $endgroup$
    – Eric Duminil
    13 hours ago




    $begingroup$
    Nice. Both are equal to $514753813311660869268255952662500831886268154$. WolframAlpha can help (1 & 2).
    $endgroup$
    – Eric Duminil
    13 hours ago











    3












    $begingroup$

    You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.



    With correct expressions you obtain:
    $$
    sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
    $$



    No contradiction appears.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.



      With correct expressions you obtain:
      $$
      sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
      $$



      No contradiction appears.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.



        With correct expressions you obtain:
        $$
        sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
        $$



        No contradiction appears.






        share|cite|improve this answer









        $endgroup$



        You made an error in applying "stars and bars". You should replace $binom{239}{40}$ with $binom{239}{39}$, and $binom{119}{20}$ with $binom{119}{19}$ and so on.



        With correct expressions you obtain:
        $$
        sum_{k=101}^{200} binom{k+19}{19}binom{219-k}{19}=frac{binom{239}{39} - binom{119}{19}^2}{2}.
        $$



        No contradiction appears.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 14 hours ago









        useruser

        5,41411030




        5,41411030






















            David ross is a new contributor. Be nice, and check out our Code of Conduct.










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            David ross is a new contributor. Be nice, and check out our Code of Conduct.













            David ross is a new contributor. Be nice, and check out our Code of Conduct.












            David ross is a new contributor. Be nice, and check out our Code of Conduct.
















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            數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
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            When can things happen in Etherscan, such as the picture below?

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            2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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            格利澤436b

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            格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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