Combinations Of Several Trials

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












Let '$n$' be the number of trials.
Let '$o$' be the number of outcomes for each trial.
Let '$a$' be the first outcome for each trial, let '$b$' be the second, '$c$' the third, etc.



How many combinations (not permutations) are there of $a, b, c, d$ (and so on) when only '$o$' and '$n$' are defined?



Let '$x$' be the number of combinations.
Let '$y$' be the number of permutations.



For example, when $o = 2$ and $n = 2$ the four permutations are: $aa, ab, ba, bb$. The number of permutations ($y$) is always $o^n$ (which here is $y = 2^2 = 4$).



But there are three combinations: two $a$'s, an $a$ and a $b$, and two $b$'s (where $ab$ and $ba$ are one combination but two permutations).



For more examples of $o$ and $n$:



  • $o=2,, n=3,, y=8,, x=4qquad (aaa, aab, abb, bbb)$

  • $o=2,, n=4,, y=16,, x=5quad:: (aaaa, aaab, aabb, abbb, bbbb)$

So when $o=2$, $y=o^n$ and $x=n+1$
But when $ogt 2$, I don't see a pattern or formula and I don't see the $^ntextC_r$ formula helping at all.



So for more examples of $o$ and $n$:



  • $o=3,, n=2,, y=9,, x=6qquad (aa, ab, ac, bb, bc, cc)$

  • $o=3,, n=3,, y=27,, x=10quad (aaa, aab, aac, abb, abc, acc, bbb, bbc, bcc, ccc)$

  • $o=3, n=4, y=81, x=15:::::: (aaaa, aaab, aaac, aabb, aabc, aacc, abbb, abbc, abcc, accc, bbbb, bbbc, bbcc, bccc, cccc)$

So it looks like when $o=3$, $x = (n+1)+(n)+(n-1)+(n-2)+...+3+2+1$. That's the sum of all natural numbers up to $n+1$ which would just simplify to $frac12(n+1)(n+2)$.



I don't know how that links to when $o=2$, but I'll continue. The $n+1$ is still there but now that $o=3$ there's an $n+2$ so I'll assume that when $o=4$ then there'll be an $n+3$. If that's the case then it would seem that each time another number is being multiplied and that new number is $o-1$ as $o$ increases. Which could be rewritten as $(n+(o-1))(n+(o-2))...(n+1) = (n+o-1)!/n!$. And since there's now a $1/2$ in front of the equation when $o=3$, I'd guess that when $o=4$, the number in front (regardless of $n$) would be either $1/3$ or $1/3!$. Which means overall the equation would either have a $1/(o-1)$ or a $1/(o-1)!$ in front of it.



So for more examples of $o$ and $n$:



  • $o=4,, n=2,, y=16,, x=10qquad (aa, ab, ac, ad, bb, bc, bd, cc, cd, dd)$

  • $o=4,, n=3,, y=64,, x=20::::::: (aaa, aab, aac, aad, abb, abc, abd, acc, acd, add, bbb, bbc, bbd, bcc, bcd, bdd, ccc, ccd, cdd, ddd)$

It looks like both of these examples apply to $(1/(o-1)!)((n+o-1)!/n!)$.



And that simplifies to $binomn+o-1n$ ... what?



So I guess my question has now changed to, can someone confirm this (that I haven't made a mistake or it doesn't apply to greater values) and why is this the case?




combinatorics combinations




share|cite|improve this question

















  • 2




    Please see the help menu for a guide on how to typeset math on this site.
    – joriki
    Jul 26 at 13:07














up vote
1
down vote

favorite












Let '$n$' be the number of trials.
Let '$o$' be the number of outcomes for each trial.
Let '$a$' be the first outcome for each trial, let '$b$' be the second, '$c$' the third, etc.



How many combinations (not permutations) are there of $a, b, c, d$ (and so on) when only '$o$' and '$n$' are defined?



Let '$x$' be the number of combinations.
Let '$y$' be the number of permutations.



For example, when $o = 2$ and $n = 2$ the four permutations are: $aa, ab, ba, bb$. The number of permutations ($y$) is always $o^n$ (which here is $y = 2^2 = 4$).



But there are three combinations: two $a$'s, an $a$ and a $b$, and two $b$'s (where $ab$ and $ba$ are one combination but two permutations).



For more examples of $o$ and $n$:



  • $o=2,, n=3,, y=8,, x=4qquad (aaa, aab, abb, bbb)$

  • $o=2,, n=4,, y=16,, x=5quad:: (aaaa, aaab, aabb, abbb, bbbb)$

So when $o=2$, $y=o^n$ and $x=n+1$
But when $ogt 2$, I don't see a pattern or formula and I don't see the $^ntextC_r$ formula helping at all.



So for more examples of $o$ and $n$:



  • $o=3,, n=2,, y=9,, x=6qquad (aa, ab, ac, bb, bc, cc)$

  • $o=3,, n=3,, y=27,, x=10quad (aaa, aab, aac, abb, abc, acc, bbb, bbc, bcc, ccc)$

  • $o=3, n=4, y=81, x=15:::::: (aaaa, aaab, aaac, aabb, aabc, aacc, abbb, abbc, abcc, accc, bbbb, bbbc, bbcc, bccc, cccc)$

So it looks like when $o=3$, $x = (n+1)+(n)+(n-1)+(n-2)+...+3+2+1$. That's the sum of all natural numbers up to $n+1$ which would just simplify to $frac12(n+1)(n+2)$.



I don't know how that links to when $o=2$, but I'll continue. The $n+1$ is still there but now that $o=3$ there's an $n+2$ so I'll assume that when $o=4$ then there'll be an $n+3$. If that's the case then it would seem that each time another number is being multiplied and that new number is $o-1$ as $o$ increases. Which could be rewritten as $(n+(o-1))(n+(o-2))...(n+1) = (n+o-1)!/n!$. And since there's now a $1/2$ in front of the equation when $o=3$, I'd guess that when $o=4$, the number in front (regardless of $n$) would be either $1/3$ or $1/3!$. Which means overall the equation would either have a $1/(o-1)$ or a $1/(o-1)!$ in front of it.



So for more examples of $o$ and $n$:



  • $o=4,, n=2,, y=16,, x=10qquad (aa, ab, ac, ad, bb, bc, bd, cc, cd, dd)$

  • $o=4,, n=3,, y=64,, x=20::::::: (aaa, aab, aac, aad, abb, abc, abd, acc, acd, add, bbb, bbc, bbd, bcc, bcd, bdd, ccc, ccd, cdd, ddd)$

It looks like both of these examples apply to $(1/(o-1)!)((n+o-1)!/n!)$.



And that simplifies to $binomn+o-1n$ ... what?



So I guess my question has now changed to, can someone confirm this (that I haven't made a mistake or it doesn't apply to greater values) and why is this the case?




combinatorics combinations




share|cite|improve this question

















  • 2




    Please see the help menu for a guide on how to typeset math on this site.
    – joriki
    Jul 26 at 13:07












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let '$n$' be the number of trials.
Let '$o$' be the number of outcomes for each trial.
Let '$a$' be the first outcome for each trial, let '$b$' be the second, '$c$' the third, etc.



How many combinations (not permutations) are there of $a, b, c, d$ (and so on) when only '$o$' and '$n$' are defined?



Let '$x$' be the number of combinations.
Let '$y$' be the number of permutations.



For example, when $o = 2$ and $n = 2$ the four permutations are: $aa, ab, ba, bb$. The number of permutations ($y$) is always $o^n$ (which here is $y = 2^2 = 4$).



But there are three combinations: two $a$'s, an $a$ and a $b$, and two $b$'s (where $ab$ and $ba$ are one combination but two permutations).



For more examples of $o$ and $n$:



  • $o=2,, n=3,, y=8,, x=4qquad (aaa, aab, abb, bbb)$

  • $o=2,, n=4,, y=16,, x=5quad:: (aaaa, aaab, aabb, abbb, bbbb)$

So when $o=2$, $y=o^n$ and $x=n+1$
But when $ogt 2$, I don't see a pattern or formula and I don't see the $^ntextC_r$ formula helping at all.



So for more examples of $o$ and $n$:



  • $o=3,, n=2,, y=9,, x=6qquad (aa, ab, ac, bb, bc, cc)$

  • $o=3,, n=3,, y=27,, x=10quad (aaa, aab, aac, abb, abc, acc, bbb, bbc, bcc, ccc)$

  • $o=3, n=4, y=81, x=15:::::: (aaaa, aaab, aaac, aabb, aabc, aacc, abbb, abbc, abcc, accc, bbbb, bbbc, bbcc, bccc, cccc)$

So it looks like when $o=3$, $x = (n+1)+(n)+(n-1)+(n-2)+...+3+2+1$. That's the sum of all natural numbers up to $n+1$ which would just simplify to $frac12(n+1)(n+2)$.



I don't know how that links to when $o=2$, but I'll continue. The $n+1$ is still there but now that $o=3$ there's an $n+2$ so I'll assume that when $o=4$ then there'll be an $n+3$. If that's the case then it would seem that each time another number is being multiplied and that new number is $o-1$ as $o$ increases. Which could be rewritten as $(n+(o-1))(n+(o-2))...(n+1) = (n+o-1)!/n!$. And since there's now a $1/2$ in front of the equation when $o=3$, I'd guess that when $o=4$, the number in front (regardless of $n$) would be either $1/3$ or $1/3!$. Which means overall the equation would either have a $1/(o-1)$ or a $1/(o-1)!$ in front of it.



So for more examples of $o$ and $n$:



  • $o=4,, n=2,, y=16,, x=10qquad (aa, ab, ac, ad, bb, bc, bd, cc, cd, dd)$

  • $o=4,, n=3,, y=64,, x=20::::::: (aaa, aab, aac, aad, abb, abc, abd, acc, acd, add, bbb, bbc, bbd, bcc, bcd, bdd, ccc, ccd, cdd, ddd)$

It looks like both of these examples apply to $(1/(o-1)!)((n+o-1)!/n!)$.



And that simplifies to $binomn+o-1n$ ... what?



So I guess my question has now changed to, can someone confirm this (that I haven't made a mistake or it doesn't apply to greater values) and why is this the case?




combinatorics combinations




share|cite|improve this question













Let '$n$' be the number of trials.
Let '$o$' be the number of outcomes for each trial.
Let '$a$' be the first outcome for each trial, let '$b$' be the second, '$c$' the third, etc.



How many combinations (not permutations) are there of $a, b, c, d$ (and so on) when only '$o$' and '$n$' are defined?



Let '$x$' be the number of combinations.
Let '$y$' be the number of permutations.



For example, when $o = 2$ and $n = 2$ the four permutations are: $aa, ab, ba, bb$. The number of permutations ($y$) is always $o^n$ (which here is $y = 2^2 = 4$).



But there are three combinations: two $a$'s, an $a$ and a $b$, and two $b$'s (where $ab$ and $ba$ are one combination but two permutations).



For more examples of $o$ and $n$:



  • $o=2,, n=3,, y=8,, x=4qquad (aaa, aab, abb, bbb)$

  • $o=2,, n=4,, y=16,, x=5quad:: (aaaa, aaab, aabb, abbb, bbbb)$

So when $o=2$, $y=o^n$ and $x=n+1$
But when $ogt 2$, I don't see a pattern or formula and I don't see the $^ntextC_r$ formula helping at all.



So for more examples of $o$ and $n$:



  • $o=3,, n=2,, y=9,, x=6qquad (aa, ab, ac, bb, bc, cc)$

  • $o=3,, n=3,, y=27,, x=10quad (aaa, aab, aac, abb, abc, acc, bbb, bbc, bcc, ccc)$

  • $o=3, n=4, y=81, x=15:::::: (aaaa, aaab, aaac, aabb, aabc, aacc, abbb, abbc, abcc, accc, bbbb, bbbc, bbcc, bccc, cccc)$

So it looks like when $o=3$, $x = (n+1)+(n)+(n-1)+(n-2)+...+3+2+1$. That's the sum of all natural numbers up to $n+1$ which would just simplify to $frac12(n+1)(n+2)$.



I don't know how that links to when $o=2$, but I'll continue. The $n+1$ is still there but now that $o=3$ there's an $n+2$ so I'll assume that when $o=4$ then there'll be an $n+3$. If that's the case then it would seem that each time another number is being multiplied and that new number is $o-1$ as $o$ increases. Which could be rewritten as $(n+(o-1))(n+(o-2))...(n+1) = (n+o-1)!/n!$. And since there's now a $1/2$ in front of the equation when $o=3$, I'd guess that when $o=4$, the number in front (regardless of $n$) would be either $1/3$ or $1/3!$. Which means overall the equation would either have a $1/(o-1)$ or a $1/(o-1)!$ in front of it.



So for more examples of $o$ and $n$:



  • $o=4,, n=2,, y=16,, x=10qquad (aa, ab, ac, ad, bb, bc, bd, cc, cd, dd)$

  • $o=4,, n=3,, y=64,, x=20::::::: (aaa, aab, aac, aad, abb, abc, abd, acc, acd, add, bbb, bbc, bbd, bcc, bcd, bdd, ccc, ccd, cdd, ddd)$

It looks like both of these examples apply to $(1/(o-1)!)((n+o-1)!/n!)$.



And that simplifies to $binomn+o-1n$ ... what?



So I guess my question has now changed to, can someone confirm this (that I haven't made a mistake or it doesn't apply to greater values) and why is this the case?





combinatorics combinations





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 13:24









N. Shales

3,0781716




3,0781716









asked Jul 26 at 12:32









Lucas Dyson-Diaz

1136




1136







  • 2




    Please see the help menu for a guide on how to typeset math on this site.
    – joriki
    Jul 26 at 13:07












  • 2




    Please see the help menu for a guide on how to typeset math on this site.
    – joriki
    Jul 26 at 13:07







2




2




Please see the help menu for a guide on how to typeset math on this site.
– joriki
Jul 26 at 13:07




Please see the help menu for a guide on how to typeset math on this site.
– joriki
Jul 26 at 13:07










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










What you have deduced is correct:



$$#(textchoices with repeats allowed)= binomn+o-1n, .$$



See this by establishing the $o$ categories (bins) separated by $o-1$ bars '$mid$' and placing $n$ balls '$bullet$' in those bins, e.g.



$$beginarrayccccccca && b && c &&cdots\
bulletbullet &mid& bullet &mid & bulletbulletbullet &mid &cdotsendarray$$



gives a combination with repeats allowed:



$$aabccccdots, .$$



Now you can see that each placement of balls in bins will yield a selection with repetition and vice versa.



Also you can see that a placement of balls in bins is just an arrangement of $o-1$ identical bars and $n$ identical balls, there are clearly



$$frac(n+o-1)!n!(o-1)!=binomn+o-1n$$



such arrangements.






share|cite|improve this answer























    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863369%2fcombinations-of-several-trials%23new-answer', 'question_page');

    );

    Post as a guest

















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    What you have deduced is correct:



    $$#(textchoices with repeats allowed)= binomn+o-1n, .$$



    See this by establishing the $o$ categories (bins) separated by $o-1$ bars '$mid$' and placing $n$ balls '$bullet$' in those bins, e.g.



    $$beginarrayccccccca && b && c &&cdots\
    bulletbullet &mid& bullet &mid & bulletbulletbullet &mid &cdotsendarray$$



    gives a combination with repeats allowed:



    $$aabccccdots, .$$



    Now you can see that each placement of balls in bins will yield a selection with repetition and vice versa.



    Also you can see that a placement of balls in bins is just an arrangement of $o-1$ identical bars and $n$ identical balls, there are clearly



    $$frac(n+o-1)!n!(o-1)!=binomn+o-1n$$



    such arrangements.






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      What you have deduced is correct:



      $$#(textchoices with repeats allowed)= binomn+o-1n, .$$



      See this by establishing the $o$ categories (bins) separated by $o-1$ bars '$mid$' and placing $n$ balls '$bullet$' in those bins, e.g.



      $$beginarrayccccccca && b && c &&cdots\
      bulletbullet &mid& bullet &mid & bulletbulletbullet &mid &cdotsendarray$$



      gives a combination with repeats allowed:



      $$aabccccdots, .$$



      Now you can see that each placement of balls in bins will yield a selection with repetition and vice versa.



      Also you can see that a placement of balls in bins is just an arrangement of $o-1$ identical bars and $n$ identical balls, there are clearly



      $$frac(n+o-1)!n!(o-1)!=binomn+o-1n$$



      such arrangements.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        What you have deduced is correct:



        $$#(textchoices with repeats allowed)= binomn+o-1n, .$$



        See this by establishing the $o$ categories (bins) separated by $o-1$ bars '$mid$' and placing $n$ balls '$bullet$' in those bins, e.g.



        $$beginarrayccccccca && b && c &&cdots\
        bulletbullet &mid& bullet &mid & bulletbulletbullet &mid &cdotsendarray$$



        gives a combination with repeats allowed:



        $$aabccccdots, .$$



        Now you can see that each placement of balls in bins will yield a selection with repetition and vice versa.



        Also you can see that a placement of balls in bins is just an arrangement of $o-1$ identical bars and $n$ identical balls, there are clearly



        $$frac(n+o-1)!n!(o-1)!=binomn+o-1n$$



        such arrangements.






        share|cite|improve this answer















        What you have deduced is correct:



        $$#(textchoices with repeats allowed)= binomn+o-1n, .$$



        See this by establishing the $o$ categories (bins) separated by $o-1$ bars '$mid$' and placing $n$ balls '$bullet$' in those bins, e.g.



        $$beginarrayccccccca && b && c &&cdots\
        bulletbullet &mid& bullet &mid & bulletbulletbullet &mid &cdotsendarray$$



        gives a combination with repeats allowed:



        $$aabccccdots, .$$



        Now you can see that each placement of balls in bins will yield a selection with repetition and vice versa.



        Also you can see that a placement of balls in bins is just an arrangement of $o-1$ identical bars and $n$ identical balls, there are clearly



        $$frac(n+o-1)!n!(o-1)!=binomn+o-1n$$



        such arrangements.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 26 at 14:46


























        answered Jul 26 at 13:05









        N. Shales

        3,0781716




        3,0781716






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863369%2fcombinations-of-several-trials%23new-answer', 'question_page');

            );

            Post as a guest
































































            Comments

            Post a Comment

            Popular posts from this blog

            What is the equation of a 3D cone with generalised tilt?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
            Read more

            Color the edges and diagonals of a regular polygon

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
            Read more

            Relationship between determinant of matrix and determinant of adjoint?

            Image
            Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
            Read more