Combinations Of Several Trials

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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




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    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

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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




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asked Jul 26 at 12:32









Lucas Dyson-Diaz

1136




1136







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










1 Answer
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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.






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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

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    up vote
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    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






















             

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