What is the probability that $m$ items drawn from $n$ distinct items contain all $n$ items?











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Suppose there are $n$ distinct items to be drawn with replacement for $m$ times, the probability of each item being drawn is assumed to be $frac{1}{n}$. What is the probability $P(m)$ that $m$ items drawn from the $n$ items contain all $n$ items?



Apparently, if $m<n$



$$P(m)=0$$



For $m=n$,



$$P(m)=P(n)=frac{n!}{n^n}$$



For $m>n$, I have no idea how to formulate the nominator.



$$P(m)=frac{?}{n^m}$$










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  • For $m>n$ I would rephrase it like the probability that no item gets drawn zero times, since you are allowed redraws of the same item.
    – M. Nestor
    Nov 15 at 23:41












  • @M.Nestor yep, it's a good problem transformation, do you have some clues about the formulation of the probability of the event "no items get drawn 0 time"?
    – Guoyang Qin
    Nov 15 at 23:47















up vote
1
down vote

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Suppose there are $n$ distinct items to be drawn with replacement for $m$ times, the probability of each item being drawn is assumed to be $frac{1}{n}$. What is the probability $P(m)$ that $m$ items drawn from the $n$ items contain all $n$ items?



Apparently, if $m<n$



$$P(m)=0$$



For $m=n$,



$$P(m)=P(n)=frac{n!}{n^n}$$



For $m>n$, I have no idea how to formulate the nominator.



$$P(m)=frac{?}{n^m}$$










share|cite|improve this question






















  • For $m>n$ I would rephrase it like the probability that no item gets drawn zero times, since you are allowed redraws of the same item.
    – M. Nestor
    Nov 15 at 23:41












  • @M.Nestor yep, it's a good problem transformation, do you have some clues about the formulation of the probability of the event "no items get drawn 0 time"?
    – Guoyang Qin
    Nov 15 at 23:47













up vote
1
down vote

favorite
1









up vote
1
down vote

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1





Suppose there are $n$ distinct items to be drawn with replacement for $m$ times, the probability of each item being drawn is assumed to be $frac{1}{n}$. What is the probability $P(m)$ that $m$ items drawn from the $n$ items contain all $n$ items?



Apparently, if $m<n$



$$P(m)=0$$



For $m=n$,



$$P(m)=P(n)=frac{n!}{n^n}$$



For $m>n$, I have no idea how to formulate the nominator.



$$P(m)=frac{?}{n^m}$$










share|cite|improve this question













Suppose there are $n$ distinct items to be drawn with replacement for $m$ times, the probability of each item being drawn is assumed to be $frac{1}{n}$. What is the probability $P(m)$ that $m$ items drawn from the $n$ items contain all $n$ items?



Apparently, if $m<n$



$$P(m)=0$$



For $m=n$,



$$P(m)=P(n)=frac{n!}{n^n}$$



For $m>n$, I have no idea how to formulate the nominator.



$$P(m)=frac{?}{n^m}$$







probability statistics sampling sampling-theory






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asked Nov 15 at 23:38









Guoyang Qin

1113




1113












  • For $m>n$ I would rephrase it like the probability that no item gets drawn zero times, since you are allowed redraws of the same item.
    – M. Nestor
    Nov 15 at 23:41












  • @M.Nestor yep, it's a good problem transformation, do you have some clues about the formulation of the probability of the event "no items get drawn 0 time"?
    – Guoyang Qin
    Nov 15 at 23:47


















  • For $m>n$ I would rephrase it like the probability that no item gets drawn zero times, since you are allowed redraws of the same item.
    – M. Nestor
    Nov 15 at 23:41












  • @M.Nestor yep, it's a good problem transformation, do you have some clues about the formulation of the probability of the event "no items get drawn 0 time"?
    – Guoyang Qin
    Nov 15 at 23:47
















For $m>n$ I would rephrase it like the probability that no item gets drawn zero times, since you are allowed redraws of the same item.
– M. Nestor
Nov 15 at 23:41






For $m>n$ I would rephrase it like the probability that no item gets drawn zero times, since you are allowed redraws of the same item.
– M. Nestor
Nov 15 at 23:41














@M.Nestor yep, it's a good problem transformation, do you have some clues about the formulation of the probability of the event "no items get drawn 0 time"?
– Guoyang Qin
Nov 15 at 23:47




@M.Nestor yep, it's a good problem transformation, do you have some clues about the formulation of the probability of the event "no items get drawn 0 time"?
– Guoyang Qin
Nov 15 at 23:47










2 Answers
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A lower bound can come from assessing the chance that a particular item has not been seen. On each draw you have $1-frac 1n$ chance of not getting it, so on $m$ draws you have $(1-frac 1n)^m$ chance of never seeing it. The chance you have not seen some item is then less than $n(1-frac 1n)^m$. This is the expected number not to see. The chance of having not seen at least one is less than this because sometimes you will have missed more than one, but if the chance of missing one is small the chance of missing two is very small. We can use the fact that $(1-frac 1n)^n approx e^{-1}$ to write the expected number to miss as $$nleft(1-frac 1nright)^m=nleft(left(1-frac 1nright)^nright)^{frac mn}approx nleft(e^{-1}right)^{frac mn}=ne^{-frac mn}$$
and the chance of missing at least one is less than this.






share|cite|improve this answer




























    up vote
    1
    down vote













    The numerator is $$n! , S_2(m,n)$$ where $S_2(m,n)$ represents a Stirling number of the second kind, sometimes written ${mbrace n}$. There is no simple closed form for these, but there are a relatively simple recurrence, generating function and inclusion-exclusion alternating sum of combinations



    For example if we call this $f(m,n)=n! , S_2(m,n)$ then we get



    $$S_2(m,n) = S_2(m-1,n-1)+n,S_2(m-1,n) \f(m,n) = n(f(m-1,n-1)+f(m-1,n))$$



    both starting with $S_2(0,0)=f(0,0)=1$ and $S_2(0,n)=f(0,n)=0$ for $n not = 0$



    For small $m$ and $n$, the values of $f(m,n)=n! , S_2(m,n)$ are



       n: 1    2    3    4    5
    m
    1 1 0 0 0 0
    2 1 2 0 0 0
    3 1 6 6 0 0
    4 1 14 36 24 0
    5 1 30 150 240 120


    and it should be fairly clear that $150=3(14+36)$ and similar patterns with the other entries






    share|cite|improve this answer























    • My understanding is that you first calculated the number of ways to split $m$ draws into $n$ partitions ($S_2(m,n)$), and then permutated the $n$ items to place ($n!$) in the $n$ partitions. Therefore, the total number is $n!S_2(m,n)$. I had a similar idea but my result for the number of ways to split is $C_{m-1}^{n-1}$, i.e., the number of ways to place $n-1$ bars into $m-1$ bins (Stars and bars)... (1/2)
      – Guoyang Qin
      Nov 16 at 3:02












    • ... I know it's incorrect, since $n!C_{m-1}^{n-1}$ would count some permutations twice. Therefore, I was wondering what the difference is between $C_{m-1}^{n-1}$ and $S_2(m,n)$? and how the Stirling number prevents itself from repeated counts? (2/2)
      – Guoyang Qin
      Nov 16 at 3:06












    • @GuoyangQin: to me personally, $n!S_2(m,n)$ is a more useful number than $S_2(m,n)$ on its own, since $S_2(m,n)$ is for distinguishable balls into indistinguishable boxes each getting at least one ball. The $n!$ comes from considering which is the first box to receive a ball ($n$ possibilities), then the second box to receive a ball ($n-1$ possibilities), and so on, i.e. making the boxes distinguishable. Stars and bars would be equivalent to counting indistinguishable balls into distinguishable boxes, and there would not be a simple translation to distinguishable balls....
      – Henry
      Nov 16 at 17:53










    • @GuoyangQin: ... In your question, you have distinguishable draws (corresponding to the balls) and distinguishable items (corresponding to the boxes).
      – Henry
      Nov 16 at 17:53










    • @GuoyangQin My $f(m,n) = n(f(m-1,n-1)+f(m-1,n))$ comes from saying that with the $m$th draw either all $n$ items have already been drawn in $f(m-1,n)$ ways so you can draw any of them this time giving $n$ possibilities, or all-except-one items have already been drawn so $n$ possibilities for the item not previously drawn and the other items have been drawn in $f(m-1,n-1)$ ways with you now needing to draw the missing item
      – Henry
      Nov 16 at 17:59













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    A lower bound can come from assessing the chance that a particular item has not been seen. On each draw you have $1-frac 1n$ chance of not getting it, so on $m$ draws you have $(1-frac 1n)^m$ chance of never seeing it. The chance you have not seen some item is then less than $n(1-frac 1n)^m$. This is the expected number not to see. The chance of having not seen at least one is less than this because sometimes you will have missed more than one, but if the chance of missing one is small the chance of missing two is very small. We can use the fact that $(1-frac 1n)^n approx e^{-1}$ to write the expected number to miss as $$nleft(1-frac 1nright)^m=nleft(left(1-frac 1nright)^nright)^{frac mn}approx nleft(e^{-1}right)^{frac mn}=ne^{-frac mn}$$
    and the chance of missing at least one is less than this.






    share|cite|improve this answer

























      up vote
      2
      down vote













      A lower bound can come from assessing the chance that a particular item has not been seen. On each draw you have $1-frac 1n$ chance of not getting it, so on $m$ draws you have $(1-frac 1n)^m$ chance of never seeing it. The chance you have not seen some item is then less than $n(1-frac 1n)^m$. This is the expected number not to see. The chance of having not seen at least one is less than this because sometimes you will have missed more than one, but if the chance of missing one is small the chance of missing two is very small. We can use the fact that $(1-frac 1n)^n approx e^{-1}$ to write the expected number to miss as $$nleft(1-frac 1nright)^m=nleft(left(1-frac 1nright)^nright)^{frac mn}approx nleft(e^{-1}right)^{frac mn}=ne^{-frac mn}$$
      and the chance of missing at least one is less than this.






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        A lower bound can come from assessing the chance that a particular item has not been seen. On each draw you have $1-frac 1n$ chance of not getting it, so on $m$ draws you have $(1-frac 1n)^m$ chance of never seeing it. The chance you have not seen some item is then less than $n(1-frac 1n)^m$. This is the expected number not to see. The chance of having not seen at least one is less than this because sometimes you will have missed more than one, but if the chance of missing one is small the chance of missing two is very small. We can use the fact that $(1-frac 1n)^n approx e^{-1}$ to write the expected number to miss as $$nleft(1-frac 1nright)^m=nleft(left(1-frac 1nright)^nright)^{frac mn}approx nleft(e^{-1}right)^{frac mn}=ne^{-frac mn}$$
        and the chance of missing at least one is less than this.






        share|cite|improve this answer












        A lower bound can come from assessing the chance that a particular item has not been seen. On each draw you have $1-frac 1n$ chance of not getting it, so on $m$ draws you have $(1-frac 1n)^m$ chance of never seeing it. The chance you have not seen some item is then less than $n(1-frac 1n)^m$. This is the expected number not to see. The chance of having not seen at least one is less than this because sometimes you will have missed more than one, but if the chance of missing one is small the chance of missing two is very small. We can use the fact that $(1-frac 1n)^n approx e^{-1}$ to write the expected number to miss as $$nleft(1-frac 1nright)^m=nleft(left(1-frac 1nright)^nright)^{frac mn}approx nleft(e^{-1}right)^{frac mn}=ne^{-frac mn}$$
        and the chance of missing at least one is less than this.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 16 at 0:14









        Ross Millikan

        288k23195365




        288k23195365






















            up vote
            1
            down vote













            The numerator is $$n! , S_2(m,n)$$ where $S_2(m,n)$ represents a Stirling number of the second kind, sometimes written ${mbrace n}$. There is no simple closed form for these, but there are a relatively simple recurrence, generating function and inclusion-exclusion alternating sum of combinations



            For example if we call this $f(m,n)=n! , S_2(m,n)$ then we get



            $$S_2(m,n) = S_2(m-1,n-1)+n,S_2(m-1,n) \f(m,n) = n(f(m-1,n-1)+f(m-1,n))$$



            both starting with $S_2(0,0)=f(0,0)=1$ and $S_2(0,n)=f(0,n)=0$ for $n not = 0$



            For small $m$ and $n$, the values of $f(m,n)=n! , S_2(m,n)$ are



               n: 1    2    3    4    5
            m
            1 1 0 0 0 0
            2 1 2 0 0 0
            3 1 6 6 0 0
            4 1 14 36 24 0
            5 1 30 150 240 120


            and it should be fairly clear that $150=3(14+36)$ and similar patterns with the other entries






            share|cite|improve this answer























            • My understanding is that you first calculated the number of ways to split $m$ draws into $n$ partitions ($S_2(m,n)$), and then permutated the $n$ items to place ($n!$) in the $n$ partitions. Therefore, the total number is $n!S_2(m,n)$. I had a similar idea but my result for the number of ways to split is $C_{m-1}^{n-1}$, i.e., the number of ways to place $n-1$ bars into $m-1$ bins (Stars and bars)... (1/2)
              – Guoyang Qin
              Nov 16 at 3:02












            • ... I know it's incorrect, since $n!C_{m-1}^{n-1}$ would count some permutations twice. Therefore, I was wondering what the difference is between $C_{m-1}^{n-1}$ and $S_2(m,n)$? and how the Stirling number prevents itself from repeated counts? (2/2)
              – Guoyang Qin
              Nov 16 at 3:06












            • @GuoyangQin: to me personally, $n!S_2(m,n)$ is a more useful number than $S_2(m,n)$ on its own, since $S_2(m,n)$ is for distinguishable balls into indistinguishable boxes each getting at least one ball. The $n!$ comes from considering which is the first box to receive a ball ($n$ possibilities), then the second box to receive a ball ($n-1$ possibilities), and so on, i.e. making the boxes distinguishable. Stars and bars would be equivalent to counting indistinguishable balls into distinguishable boxes, and there would not be a simple translation to distinguishable balls....
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin: ... In your question, you have distinguishable draws (corresponding to the balls) and distinguishable items (corresponding to the boxes).
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin My $f(m,n) = n(f(m-1,n-1)+f(m-1,n))$ comes from saying that with the $m$th draw either all $n$ items have already been drawn in $f(m-1,n)$ ways so you can draw any of them this time giving $n$ possibilities, or all-except-one items have already been drawn so $n$ possibilities for the item not previously drawn and the other items have been drawn in $f(m-1,n-1)$ ways with you now needing to draw the missing item
              – Henry
              Nov 16 at 17:59

















            up vote
            1
            down vote













            The numerator is $$n! , S_2(m,n)$$ where $S_2(m,n)$ represents a Stirling number of the second kind, sometimes written ${mbrace n}$. There is no simple closed form for these, but there are a relatively simple recurrence, generating function and inclusion-exclusion alternating sum of combinations



            For example if we call this $f(m,n)=n! , S_2(m,n)$ then we get



            $$S_2(m,n) = S_2(m-1,n-1)+n,S_2(m-1,n) \f(m,n) = n(f(m-1,n-1)+f(m-1,n))$$



            both starting with $S_2(0,0)=f(0,0)=1$ and $S_2(0,n)=f(0,n)=0$ for $n not = 0$



            For small $m$ and $n$, the values of $f(m,n)=n! , S_2(m,n)$ are



               n: 1    2    3    4    5
            m
            1 1 0 0 0 0
            2 1 2 0 0 0
            3 1 6 6 0 0
            4 1 14 36 24 0
            5 1 30 150 240 120


            and it should be fairly clear that $150=3(14+36)$ and similar patterns with the other entries






            share|cite|improve this answer























            • My understanding is that you first calculated the number of ways to split $m$ draws into $n$ partitions ($S_2(m,n)$), and then permutated the $n$ items to place ($n!$) in the $n$ partitions. Therefore, the total number is $n!S_2(m,n)$. I had a similar idea but my result for the number of ways to split is $C_{m-1}^{n-1}$, i.e., the number of ways to place $n-1$ bars into $m-1$ bins (Stars and bars)... (1/2)
              – Guoyang Qin
              Nov 16 at 3:02












            • ... I know it's incorrect, since $n!C_{m-1}^{n-1}$ would count some permutations twice. Therefore, I was wondering what the difference is between $C_{m-1}^{n-1}$ and $S_2(m,n)$? and how the Stirling number prevents itself from repeated counts? (2/2)
              – Guoyang Qin
              Nov 16 at 3:06












            • @GuoyangQin: to me personally, $n!S_2(m,n)$ is a more useful number than $S_2(m,n)$ on its own, since $S_2(m,n)$ is for distinguishable balls into indistinguishable boxes each getting at least one ball. The $n!$ comes from considering which is the first box to receive a ball ($n$ possibilities), then the second box to receive a ball ($n-1$ possibilities), and so on, i.e. making the boxes distinguishable. Stars and bars would be equivalent to counting indistinguishable balls into distinguishable boxes, and there would not be a simple translation to distinguishable balls....
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin: ... In your question, you have distinguishable draws (corresponding to the balls) and distinguishable items (corresponding to the boxes).
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin My $f(m,n) = n(f(m-1,n-1)+f(m-1,n))$ comes from saying that with the $m$th draw either all $n$ items have already been drawn in $f(m-1,n)$ ways so you can draw any of them this time giving $n$ possibilities, or all-except-one items have already been drawn so $n$ possibilities for the item not previously drawn and the other items have been drawn in $f(m-1,n-1)$ ways with you now needing to draw the missing item
              – Henry
              Nov 16 at 17:59















            up vote
            1
            down vote










            up vote
            1
            down vote









            The numerator is $$n! , S_2(m,n)$$ where $S_2(m,n)$ represents a Stirling number of the second kind, sometimes written ${mbrace n}$. There is no simple closed form for these, but there are a relatively simple recurrence, generating function and inclusion-exclusion alternating sum of combinations



            For example if we call this $f(m,n)=n! , S_2(m,n)$ then we get



            $$S_2(m,n) = S_2(m-1,n-1)+n,S_2(m-1,n) \f(m,n) = n(f(m-1,n-1)+f(m-1,n))$$



            both starting with $S_2(0,0)=f(0,0)=1$ and $S_2(0,n)=f(0,n)=0$ for $n not = 0$



            For small $m$ and $n$, the values of $f(m,n)=n! , S_2(m,n)$ are



               n: 1    2    3    4    5
            m
            1 1 0 0 0 0
            2 1 2 0 0 0
            3 1 6 6 0 0
            4 1 14 36 24 0
            5 1 30 150 240 120


            and it should be fairly clear that $150=3(14+36)$ and similar patterns with the other entries






            share|cite|improve this answer














            The numerator is $$n! , S_2(m,n)$$ where $S_2(m,n)$ represents a Stirling number of the second kind, sometimes written ${mbrace n}$. There is no simple closed form for these, but there are a relatively simple recurrence, generating function and inclusion-exclusion alternating sum of combinations



            For example if we call this $f(m,n)=n! , S_2(m,n)$ then we get



            $$S_2(m,n) = S_2(m-1,n-1)+n,S_2(m-1,n) \f(m,n) = n(f(m-1,n-1)+f(m-1,n))$$



            both starting with $S_2(0,0)=f(0,0)=1$ and $S_2(0,n)=f(0,n)=0$ for $n not = 0$



            For small $m$ and $n$, the values of $f(m,n)=n! , S_2(m,n)$ are



               n: 1    2    3    4    5
            m
            1 1 0 0 0 0
            2 1 2 0 0 0
            3 1 6 6 0 0
            4 1 14 36 24 0
            5 1 30 150 240 120


            and it should be fairly clear that $150=3(14+36)$ and similar patterns with the other entries







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 16 at 1:02

























            answered Nov 16 at 0:51









            Henry

            96.9k474154




            96.9k474154












            • My understanding is that you first calculated the number of ways to split $m$ draws into $n$ partitions ($S_2(m,n)$), and then permutated the $n$ items to place ($n!$) in the $n$ partitions. Therefore, the total number is $n!S_2(m,n)$. I had a similar idea but my result for the number of ways to split is $C_{m-1}^{n-1}$, i.e., the number of ways to place $n-1$ bars into $m-1$ bins (Stars and bars)... (1/2)
              – Guoyang Qin
              Nov 16 at 3:02












            • ... I know it's incorrect, since $n!C_{m-1}^{n-1}$ would count some permutations twice. Therefore, I was wondering what the difference is between $C_{m-1}^{n-1}$ and $S_2(m,n)$? and how the Stirling number prevents itself from repeated counts? (2/2)
              – Guoyang Qin
              Nov 16 at 3:06












            • @GuoyangQin: to me personally, $n!S_2(m,n)$ is a more useful number than $S_2(m,n)$ on its own, since $S_2(m,n)$ is for distinguishable balls into indistinguishable boxes each getting at least one ball. The $n!$ comes from considering which is the first box to receive a ball ($n$ possibilities), then the second box to receive a ball ($n-1$ possibilities), and so on, i.e. making the boxes distinguishable. Stars and bars would be equivalent to counting indistinguishable balls into distinguishable boxes, and there would not be a simple translation to distinguishable balls....
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin: ... In your question, you have distinguishable draws (corresponding to the balls) and distinguishable items (corresponding to the boxes).
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin My $f(m,n) = n(f(m-1,n-1)+f(m-1,n))$ comes from saying that with the $m$th draw either all $n$ items have already been drawn in $f(m-1,n)$ ways so you can draw any of them this time giving $n$ possibilities, or all-except-one items have already been drawn so $n$ possibilities for the item not previously drawn and the other items have been drawn in $f(m-1,n-1)$ ways with you now needing to draw the missing item
              – Henry
              Nov 16 at 17:59




















            • My understanding is that you first calculated the number of ways to split $m$ draws into $n$ partitions ($S_2(m,n)$), and then permutated the $n$ items to place ($n!$) in the $n$ partitions. Therefore, the total number is $n!S_2(m,n)$. I had a similar idea but my result for the number of ways to split is $C_{m-1}^{n-1}$, i.e., the number of ways to place $n-1$ bars into $m-1$ bins (Stars and bars)... (1/2)
              – Guoyang Qin
              Nov 16 at 3:02












            • ... I know it's incorrect, since $n!C_{m-1}^{n-1}$ would count some permutations twice. Therefore, I was wondering what the difference is between $C_{m-1}^{n-1}$ and $S_2(m,n)$? and how the Stirling number prevents itself from repeated counts? (2/2)
              – Guoyang Qin
              Nov 16 at 3:06












            • @GuoyangQin: to me personally, $n!S_2(m,n)$ is a more useful number than $S_2(m,n)$ on its own, since $S_2(m,n)$ is for distinguishable balls into indistinguishable boxes each getting at least one ball. The $n!$ comes from considering which is the first box to receive a ball ($n$ possibilities), then the second box to receive a ball ($n-1$ possibilities), and so on, i.e. making the boxes distinguishable. Stars and bars would be equivalent to counting indistinguishable balls into distinguishable boxes, and there would not be a simple translation to distinguishable balls....
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin: ... In your question, you have distinguishable draws (corresponding to the balls) and distinguishable items (corresponding to the boxes).
              – Henry
              Nov 16 at 17:53










            • @GuoyangQin My $f(m,n) = n(f(m-1,n-1)+f(m-1,n))$ comes from saying that with the $m$th draw either all $n$ items have already been drawn in $f(m-1,n)$ ways so you can draw any of them this time giving $n$ possibilities, or all-except-one items have already been drawn so $n$ possibilities for the item not previously drawn and the other items have been drawn in $f(m-1,n-1)$ ways with you now needing to draw the missing item
              – Henry
              Nov 16 at 17:59


















            My understanding is that you first calculated the number of ways to split $m$ draws into $n$ partitions ($S_2(m,n)$), and then permutated the $n$ items to place ($n!$) in the $n$ partitions. Therefore, the total number is $n!S_2(m,n)$. I had a similar idea but my result for the number of ways to split is $C_{m-1}^{n-1}$, i.e., the number of ways to place $n-1$ bars into $m-1$ bins (Stars and bars)... (1/2)
            – Guoyang Qin
            Nov 16 at 3:02






            My understanding is that you first calculated the number of ways to split $m$ draws into $n$ partitions ($S_2(m,n)$), and then permutated the $n$ items to place ($n!$) in the $n$ partitions. Therefore, the total number is $n!S_2(m,n)$. I had a similar idea but my result for the number of ways to split is $C_{m-1}^{n-1}$, i.e., the number of ways to place $n-1$ bars into $m-1$ bins (Stars and bars)... (1/2)
            – Guoyang Qin
            Nov 16 at 3:02














            ... I know it's incorrect, since $n!C_{m-1}^{n-1}$ would count some permutations twice. Therefore, I was wondering what the difference is between $C_{m-1}^{n-1}$ and $S_2(m,n)$? and how the Stirling number prevents itself from repeated counts? (2/2)
            – Guoyang Qin
            Nov 16 at 3:06






            ... I know it's incorrect, since $n!C_{m-1}^{n-1}$ would count some permutations twice. Therefore, I was wondering what the difference is between $C_{m-1}^{n-1}$ and $S_2(m,n)$? and how the Stirling number prevents itself from repeated counts? (2/2)
            – Guoyang Qin
            Nov 16 at 3:06














            @GuoyangQin: to me personally, $n!S_2(m,n)$ is a more useful number than $S_2(m,n)$ on its own, since $S_2(m,n)$ is for distinguishable balls into indistinguishable boxes each getting at least one ball. The $n!$ comes from considering which is the first box to receive a ball ($n$ possibilities), then the second box to receive a ball ($n-1$ possibilities), and so on, i.e. making the boxes distinguishable. Stars and bars would be equivalent to counting indistinguishable balls into distinguishable boxes, and there would not be a simple translation to distinguishable balls....
            – Henry
            Nov 16 at 17:53




            @GuoyangQin: to me personally, $n!S_2(m,n)$ is a more useful number than $S_2(m,n)$ on its own, since $S_2(m,n)$ is for distinguishable balls into indistinguishable boxes each getting at least one ball. The $n!$ comes from considering which is the first box to receive a ball ($n$ possibilities), then the second box to receive a ball ($n-1$ possibilities), and so on, i.e. making the boxes distinguishable. Stars and bars would be equivalent to counting indistinguishable balls into distinguishable boxes, and there would not be a simple translation to distinguishable balls....
            – Henry
            Nov 16 at 17:53












            @GuoyangQin: ... In your question, you have distinguishable draws (corresponding to the balls) and distinguishable items (corresponding to the boxes).
            – Henry
            Nov 16 at 17:53




            @GuoyangQin: ... In your question, you have distinguishable draws (corresponding to the balls) and distinguishable items (corresponding to the boxes).
            – Henry
            Nov 16 at 17:53












            @GuoyangQin My $f(m,n) = n(f(m-1,n-1)+f(m-1,n))$ comes from saying that with the $m$th draw either all $n$ items have already been drawn in $f(m-1,n)$ ways so you can draw any of them this time giving $n$ possibilities, or all-except-one items have already been drawn so $n$ possibilities for the item not previously drawn and the other items have been drawn in $f(m-1,n-1)$ ways with you now needing to draw the missing item
            – Henry
            Nov 16 at 17:59






            @GuoyangQin My $f(m,n) = n(f(m-1,n-1)+f(m-1,n))$ comes from saying that with the $m$th draw either all $n$ items have already been drawn in $f(m-1,n)$ ways so you can draw any of them this time giving $n$ possibilities, or all-except-one items have already been drawn so $n$ possibilities for the item not previously drawn and the other items have been drawn in $f(m-1,n-1)$ ways with you now needing to draw the missing item
            – Henry
            Nov 16 at 17:59




















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