Selecting n<m elements from a set of m elements containing repeated elements, What's the SIZE of the...












0












$begingroup$


For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?



I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.






combinatorics







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    0












    $begingroup$


    For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?



    I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.






    combinatorics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?



      I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.






      combinatorics







      share|cite|improve this question











      $endgroup$




      For example, there's $21$ objects in a hat, $7$ pink balls, $7$ purple balls, $7$ yellow balls. You only pick $5$ WITHOUT REPLACEMENT. What is the SIZE of the sample space?



      I see some people saying its $C(21,5)$ but that doesn't make sense to me because of repeated elements and the overcounting that happens when you just simply do $C(21,5)$.






      combinatorics




      combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 9:27









      Tianlalu

      3,08621038




      3,08621038










      asked Dec 3 '18 at 2:17









      KrioKrio

      306




      306






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".



          The first is just $C(21,5)$, but as you noted not all of them are distinguishable.



          To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.



          This is $C(5+2,2)$ with the following stars and bars argument:




          • Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"

          • Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow

          • Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5

          • The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
            $endgroup$
            – Krio
            Dec 3 '18 at 2:39








          • 1




            $begingroup$
            If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:43












          • $begingroup$
            thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
            $endgroup$
            – Krio
            Dec 3 '18 at 2:52






          • 1




            $begingroup$
            When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:56



















          0












          $begingroup$

          It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
            $endgroup$
            – Krio
            Dec 3 '18 at 2:35










          • $begingroup$
            Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
            $endgroup$
            – Boshu
            Dec 3 '18 at 2:37











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".



          The first is just $C(21,5)$, but as you noted not all of them are distinguishable.



          To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.



          This is $C(5+2,2)$ with the following stars and bars argument:




          • Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"

          • Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow

          • Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5

          • The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
            $endgroup$
            – Krio
            Dec 3 '18 at 2:39








          • 1




            $begingroup$
            If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:43












          • $begingroup$
            thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
            $endgroup$
            – Krio
            Dec 3 '18 at 2:52






          • 1




            $begingroup$
            When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:56
















          1












          $begingroup$

          This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".



          The first is just $C(21,5)$, but as you noted not all of them are distinguishable.



          To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.



          This is $C(5+2,2)$ with the following stars and bars argument:




          • Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"

          • Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow

          • Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5

          • The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
            $endgroup$
            – Krio
            Dec 3 '18 at 2:39








          • 1




            $begingroup$
            If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:43












          • $begingroup$
            thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
            $endgroup$
            – Krio
            Dec 3 '18 at 2:52






          • 1




            $begingroup$
            When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:56














          1












          1








          1





          $begingroup$

          This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".



          The first is just $C(21,5)$, but as you noted not all of them are distinguishable.



          To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.



          This is $C(5+2,2)$ with the following stars and bars argument:




          • Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"

          • Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow

          • Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5

          • The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.






          share|cite|improve this answer









          $endgroup$



          This depends on whether you take "size" to mean "number of equally probable events" or "number of distinguishable events".



          The first is just $C(21,5)$, but as you noted not all of them are distinguishable.



          To count the number of distinguishable events, just think about what you can actually tell from your sample: (number of pink balls, number of purple balls, number of yellow balls). Any number can be 0-5 inclusive and they sum to exactly 5.



          This is $C(5+2,2)$ with the following stars and bars argument:




          • Place seven items, 5 indistinguishable balls of unspecified color and 2 "dividers"

          • Call all balls in front of both dividers pink, balls in between purple, and balls behind both yellow

          • Each possible placement now is a distinguishable (#pink, #purple, #yellow) arrangement such that the numbers sum to exactly 5

          • The number of placements is therefore $C(5+2,2)$ by choosing the divider positions.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 2:33









          obscuransobscurans

          1,152311




          1,152311












          • $begingroup$
            This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
            $endgroup$
            – Krio
            Dec 3 '18 at 2:39








          • 1




            $begingroup$
            If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:43












          • $begingroup$
            thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
            $endgroup$
            – Krio
            Dec 3 '18 at 2:52






          • 1




            $begingroup$
            When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:56


















          • $begingroup$
            This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
            $endgroup$
            – Krio
            Dec 3 '18 at 2:39








          • 1




            $begingroup$
            If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:43












          • $begingroup$
            thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
            $endgroup$
            – Krio
            Dec 3 '18 at 2:52






          • 1




            $begingroup$
            When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
            $endgroup$
            – obscurans
            Dec 3 '18 at 2:56
















          $begingroup$
          This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
          $endgroup$
          – Krio
          Dec 3 '18 at 2:39






          $begingroup$
          This makes sense to me, to give context I'm trying to calculate the probability of picking 4 of the same colour and 1 different colour when picking 5. In this scenario I'm confused which sample space size I should be using
          $endgroup$
          – Krio
          Dec 3 '18 at 2:39






          1




          1




          $begingroup$
          If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
          $endgroup$
          – obscurans
          Dec 3 '18 at 2:43






          $begingroup$
          If you want to divide by a denominator, you definitely want to use "number of equally probable events", which is just $C(21,5)$.
          $endgroup$
          – obscurans
          Dec 3 '18 at 2:43














          $begingroup$
          thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
          $endgroup$
          – Krio
          Dec 3 '18 at 2:52




          $begingroup$
          thank you. Sort of makes sense to me but i still have some things that are conflicting with me. Going to ponder on it a bit more
          $endgroup$
          – Krio
          Dec 3 '18 at 2:52




          1




          1




          $begingroup$
          When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
          $endgroup$
          – obscurans
          Dec 3 '18 at 2:56




          $begingroup$
          When you calculate probability(pick 4 same colour) = (ways to pick 4 same colour)/(ways to pick 5 balls), what you're doing is saying "each of the ways to pick 5 balls is equally likely, so their probability is 1/(#ways)"
          $endgroup$
          – obscurans
          Dec 3 '18 at 2:56











          0












          $begingroup$

          It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
            $endgroup$
            – Krio
            Dec 3 '18 at 2:35










          • $begingroup$
            Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
            $endgroup$
            – Boshu
            Dec 3 '18 at 2:37
















          0












          $begingroup$

          It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
            $endgroup$
            – Krio
            Dec 3 '18 at 2:35










          • $begingroup$
            Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
            $endgroup$
            – Boshu
            Dec 3 '18 at 2:37














          0












          0








          0





          $begingroup$

          It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.






          share|cite|improve this answer









          $endgroup$



          It should be $21 choose 5$ because even though the balls look the same, they are not the same, so if you have the balls ${p_1,ldots,p_7}$,${r_1,ldots,r_7}$ and ${y_1,ldots,y_7}$ and you're supposed to pick $3$ of them, picking $p_1,r_1,y_1$ is not the same event where you pick $p_5,r_5,y_5$. You might want to look up Maxwell Boltzmann statistics and Bose Einstein statistics.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 2:29









          BoshuBoshu

          703315




          703315












          • $begingroup$
            for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
            $endgroup$
            – Krio
            Dec 3 '18 at 2:35










          • $begingroup$
            Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
            $endgroup$
            – Boshu
            Dec 3 '18 at 2:37


















          • $begingroup$
            for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
            $endgroup$
            – Krio
            Dec 3 '18 at 2:35










          • $begingroup$
            Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
            $endgroup$
            – Boshu
            Dec 3 '18 at 2:37
















          $begingroup$
          for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
          $endgroup$
          – Krio
          Dec 3 '18 at 2:35




          $begingroup$
          for context, I'm trying to use that number in the denominator of a probability calculation. So do repeated instances as you pointed out count as different events in the sample space?
          $endgroup$
          – Krio
          Dec 3 '18 at 2:35












          $begingroup$
          Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
          $endgroup$
          – Boshu
          Dec 3 '18 at 2:37




          $begingroup$
          Ideally yes. It depends on what you define as distinguishable. For all classical particles, you're supposed to count it as such. Check the last sentence of my answer and look it up. Might be of help to you.
          $endgroup$
          – Boshu
          Dec 3 '18 at 2:37


















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