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












5












$begingroup$



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 possiblities that there are more balls in the first $20$ jars is equal to the number of possiblities 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 come out to $sum_{k=101}^{200} binom{k+19}{20}binom{219-k}{20}$.






combinatorics







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New contributor




David ross is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    2 hours ago










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










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












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










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
















5












$begingroup$



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 possiblities that there are more balls in the first $20$ jars is equal to the number of possiblities 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 come 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
    2 hours ago










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










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












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










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














5












5








5


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 possiblities that there are more balls in the first $20$ jars is equal to the number of possiblities 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 come 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 possiblities that there are more balls in the first $20$ jars is equal to the number of possiblities 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 come 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 20 mins 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 2 hours ago









David rossDavid ross

261




261




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
    2 hours ago










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










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












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










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










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










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












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










  • $begingroup$
    @user Still waking up. Yes, you are correct.
    $endgroup$
    – N. F. Taussig
    1 hour 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
2 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
2 hours ago












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




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












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






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














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




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












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




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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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
    6 mins ago



















1












$begingroup$

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









    2












    $begingroup$

    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
      6 mins ago
















    2












    $begingroup$

    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
      6 mins ago














    2












    2








    2





    $begingroup$

    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 1 hour ago









    drhabdrhab

    103k545136




    103k545136












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


















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
















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




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











    1












    $begingroup$

    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









    $endgroup$


















      1












      $begingroup$

      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









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 1 hour ago









        useruser

        5,38211030




        5,38211030






















            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.













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
















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            How to solve the problem of ntp “Unable to contact time server” from KDE?

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            1 0 I want to set date and time automatically but this error appears when I try to .. I'm using openSUSE, I tried different servers but it didn't work either, help please! It doesn't seem to be a network problem: $ ping pool.ntp.org 64 bytes from 41.78.128.17: icmp_seq=1 ttl=41 time=260 ms To the question of whether ntpd is installed and running: $ ps -C ntpd PID TTY TIME CMD i.e. there's no ntpd process running. $ rpm -qa | grep ntp yast2-ntp-client-3.1.12-1.7.noarch ntp-4.2.6p5-25.2.1.i586 Further information from the comments: $ sudo ntpdate pool.ntp.org 25 Sep 19:41:51 ntpdate[3162]: no server suitable for synchronization found networking kde opensuse ntp ntpd sh...
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            ASUS Zenbook UX433/UX333 — Configure Touchpad-embedded numpad on Linux

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            0 1 I have an ASUS Zenbook UX433,on which ASUS decided to innovate by embedding a Numpad inside the touchpad (see picture 1). Toggling between classic touchpad mode and numpad (also turns the touchpad LEDs on) can be done through the golden button in the top-right corner of the touchpad. However, on my Manjaro arch Linux config, this behavior does not work by default and I can only use the touchpad, no numpad. Has anyone been able to configure the numpad to work on Linux, or have any idea how this could be achieved? EDIT1: I have had a look at the official Windows drivers for the UX433FA TouchPad, but was not able to make anything of it. I tried installing them on Linux using tools like NDISWrapper, but they do not detect the TouchPad/Numpad hardware, and my understanding of drivers is too limited. ...
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