Question about permutation?












0












$begingroup$


Can someone help me with this permutation exercises...



1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..



2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



This one is also without replacement, so it would be



$P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.



Your help will be appreciated.






permutations







share|cite|improve this question









$endgroup$












  • $begingroup$
    1) is correct, assumingone distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats. For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?
    $endgroup$
    – Bram28
    Dec 14 '18 at 16:52


















0












$begingroup$


Can someone help me with this permutation exercises...



1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..



2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



This one is also without replacement, so it would be



$P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.



Your help will be appreciated.






permutations







share|cite|improve this question









$endgroup$












  • $begingroup$
    1) is correct, assumingone distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats. For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?
    $endgroup$
    – Bram28
    Dec 14 '18 at 16:52
















0












0








0





$begingroup$


Can someone help me with this permutation exercises...



1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..



2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



This one is also without replacement, so it would be



$P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.



Your help will be appreciated.






permutations







share|cite|improve this question









$endgroup$




Can someone help me with this permutation exercises...



1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..



2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



This one is also without replacement, so it would be



$P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.



Your help will be appreciated.






permutations




permutations






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share|cite|improve this question











share|cite|improve this question




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asked Dec 14 '18 at 16:46









gi2302gi2302

103




103












  • $begingroup$
    1) is correct, assumingone distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats. For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?
    $endgroup$
    – Bram28
    Dec 14 '18 at 16:52




















  • $begingroup$
    1) is correct, assumingone distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats. For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?
    $endgroup$
    – Bram28
    Dec 14 '18 at 16:52


















$begingroup$
1) is correct, assumingone distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats. For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?
$endgroup$
– Bram28
Dec 14 '18 at 16:52






$begingroup$
1) is correct, assumingone distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats. For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?
$endgroup$
– Bram28
Dec 14 '18 at 16:52












4 Answers
4






active

oldest

votes


















2












$begingroup$

Your first is correct if you regard the floats and orchestras as distinct. I would read the problem to consider all the floats as interchangeable and the orchestras likewise. Now you just need to choose three places in line to put orchestras.



In the second the dogs are clearly distinguishable and we assume the people are, too. That makes your answer correct.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hints



    For (1) you are assuming the floats and the orchestras are all different. If that were the case then your answer of $11!$ is right. But the question suggests that the orchestras are interchangeable, as are the floats. So find the number of positions for the orchestras.



    (In the future, one question per post please.)






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$


      1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



      Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..




      This is alright if all the floats and orchestras are distinguishable. If all 11 'objects' are, then you can permute them in $11!$ ways. I assume they mean that all the floats aren't distinguishable, the same for the orchestras. In that case you have to compensate for the fact that permuting the floats ($8!$ ways) and the orchestras ($3!$ ways) don't change the line-up:
      $$frac{11!}{8!;3!} = ldots$$




      2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



      This one is also without replacement, so it would be



      $P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.




      Looks good, unless a person can win more than one prize (dog)...?






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I'm assuming, a person can't win more than one dog.
        $endgroup$
        – gi2302
        Dec 14 '18 at 17:33










      • $begingroup$
        What u mean with distinguishable...?
        $endgroup$
        – gi2302
        Dec 14 '18 at 17:35










      • $begingroup$
        Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised.
        $endgroup$
        – StackTD
        Dec 14 '18 at 17:35










      • $begingroup$
        Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!).
        $endgroup$
        – StackTD
        Dec 14 '18 at 17:36





















      0












      $begingroup$

      1) is correct ... assuming one distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other ... personally, I couldn't take two orchestras in a row ...), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats.



      For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?






      share|cite|improve this answer









      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Your first is correct if you regard the floats and orchestras as distinct. I would read the problem to consider all the floats as interchangeable and the orchestras likewise. Now you just need to choose three places in line to put orchestras.



        In the second the dogs are clearly distinguishable and we assume the people are, too. That makes your answer correct.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Your first is correct if you regard the floats and orchestras as distinct. I would read the problem to consider all the floats as interchangeable and the orchestras likewise. Now you just need to choose three places in line to put orchestras.



          In the second the dogs are clearly distinguishable and we assume the people are, too. That makes your answer correct.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Your first is correct if you regard the floats and orchestras as distinct. I would read the problem to consider all the floats as interchangeable and the orchestras likewise. Now you just need to choose three places in line to put orchestras.



            In the second the dogs are clearly distinguishable and we assume the people are, too. That makes your answer correct.






            share|cite|improve this answer









            $endgroup$



            Your first is correct if you regard the floats and orchestras as distinct. I would read the problem to consider all the floats as interchangeable and the orchestras likewise. Now you just need to choose three places in line to put orchestras.



            In the second the dogs are clearly distinguishable and we assume the people are, too. That makes your answer correct.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 14 '18 at 16:51









            Ross MillikanRoss Millikan

            295k23198371




            295k23198371























                1












                $begingroup$

                Hints



                For (1) you are assuming the floats and the orchestras are all different. If that were the case then your answer of $11!$ is right. But the question suggests that the orchestras are interchangeable, as are the floats. So find the number of positions for the orchestras.



                (In the future, one question per post please.)






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Hints



                  For (1) you are assuming the floats and the orchestras are all different. If that were the case then your answer of $11!$ is right. But the question suggests that the orchestras are interchangeable, as are the floats. So find the number of positions for the orchestras.



                  (In the future, one question per post please.)






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Hints



                    For (1) you are assuming the floats and the orchestras are all different. If that were the case then your answer of $11!$ is right. But the question suggests that the orchestras are interchangeable, as are the floats. So find the number of positions for the orchestras.



                    (In the future, one question per post please.)






                    share|cite|improve this answer









                    $endgroup$



                    Hints



                    For (1) you are assuming the floats and the orchestras are all different. If that were the case then your answer of $11!$ is right. But the question suggests that the orchestras are interchangeable, as are the floats. So find the number of positions for the orchestras.



                    (In the future, one question per post please.)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 14 '18 at 16:52









                    Ethan BolkerEthan Bolker

                    42.7k549113




                    42.7k549113























                        1












                        $begingroup$


                        1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



                        Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..




                        This is alright if all the floats and orchestras are distinguishable. If all 11 'objects' are, then you can permute them in $11!$ ways. I assume they mean that all the floats aren't distinguishable, the same for the orchestras. In that case you have to compensate for the fact that permuting the floats ($8!$ ways) and the orchestras ($3!$ ways) don't change the line-up:
                        $$frac{11!}{8!;3!} = ldots$$




                        2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



                        This one is also without replacement, so it would be



                        $P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.




                        Looks good, unless a person can win more than one prize (dog)...?






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          I'm assuming, a person can't win more than one dog.
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:33










                        • $begingroup$
                          What u mean with distinguishable...?
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised.
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!).
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:36


















                        1












                        $begingroup$


                        1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



                        Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..




                        This is alright if all the floats and orchestras are distinguishable. If all 11 'objects' are, then you can permute them in $11!$ ways. I assume they mean that all the floats aren't distinguishable, the same for the orchestras. In that case you have to compensate for the fact that permuting the floats ($8!$ ways) and the orchestras ($3!$ ways) don't change the line-up:
                        $$frac{11!}{8!;3!} = ldots$$




                        2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



                        This one is also without replacement, so it would be



                        $P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.




                        Looks good, unless a person can win more than one prize (dog)...?






                        share|cite|improve this answer









                        $endgroup$













                        • $begingroup$
                          I'm assuming, a person can't win more than one dog.
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:33










                        • $begingroup$
                          What u mean with distinguishable...?
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised.
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!).
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:36
















                        1












                        1








                        1





                        $begingroup$


                        1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



                        Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..




                        This is alright if all the floats and orchestras are distinguishable. If all 11 'objects' are, then you can permute them in $11!$ ways. I assume they mean that all the floats aren't distinguishable, the same for the orchestras. In that case you have to compensate for the fact that permuting the floats ($8!$ ways) and the orchestras ($3!$ ways) don't change the line-up:
                        $$frac{11!}{8!;3!} = ldots$$




                        2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



                        This one is also without replacement, so it would be



                        $P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.




                        Looks good, unless a person can win more than one prize (dog)...?






                        share|cite|improve this answer









                        $endgroup$




                        1) In a NY parade, there are 8 floats and 3 orchestras. How many possible orders are there?



                        Ok. Tell me if I'm right. There is no repetition in here. So if I'm going to line them up this will be a $P(11,11)=frac{11!}{(11-11)!}=frac{11!}{0!}=frac{11!}{1}=39,916,800$. But I don't know this is a huge number. So I don't know if make sense at all..




                        This is alright if all the floats and orchestras are distinguishable. If all 11 'objects' are, then you can permute them in $11!$ ways. I assume they mean that all the floats aren't distinguishable, the same for the orchestras. In that case you have to compensate for the fact that permuting the floats ($8!$ ways) and the orchestras ($3!$ ways) don't change the line-up:
                        $$frac{11!}{8!;3!} = ldots$$




                        2) A Labrador Retriever, a Siberian Husky, and a German Sheperd are raffled at a pet store. If there are 10 participants, how many ways can the prizes be allocated?



                        This one is also without replacement, so it would be



                        $P(10,3)=frac{10!}{(10-3)!}=frac{10!}{7!}=10(9)(8)=720$.




                        Looks good, unless a person can win more than one prize (dog)...?







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 14 '18 at 16:52









                        StackTDStackTD

                        22.7k2050




                        22.7k2050












                        • $begingroup$
                          I'm assuming, a person can't win more than one dog.
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:33










                        • $begingroup$
                          What u mean with distinguishable...?
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised.
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!).
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:36




















                        • $begingroup$
                          I'm assuming, a person can't win more than one dog.
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:33










                        • $begingroup$
                          What u mean with distinguishable...?
                          $endgroup$
                          – gi2302
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised.
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:35










                        • $begingroup$
                          Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!).
                          $endgroup$
                          – StackTD
                          Dec 14 '18 at 17:36


















                        $begingroup$
                        I'm assuming, a person can't win more than one dog.
                        $endgroup$
                        – gi2302
                        Dec 14 '18 at 17:33




                        $begingroup$
                        I'm assuming, a person can't win more than one dog.
                        $endgroup$
                        – gi2302
                        Dec 14 '18 at 17:33












                        $begingroup$
                        What u mean with distinguishable...?
                        $endgroup$
                        – gi2302
                        Dec 14 '18 at 17:35




                        $begingroup$
                        What u mean with distinguishable...?
                        $endgroup$
                        – gi2302
                        Dec 14 '18 at 17:35












                        $begingroup$
                        Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised.
                        $endgroup$
                        – StackTD
                        Dec 14 '18 at 17:35




                        $begingroup$
                        Q2: Then you're right, but note that it makes a difference and that's it not that obvious that this is the correct interpretation: one person could have three winning raffle tickets, for example! You can't really tell without knowing how the raffle is organised.
                        $endgroup$
                        – StackTD
                        Dec 14 '18 at 17:35












                        $begingroup$
                        Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!).
                        $endgroup$
                        – StackTD
                        Dec 14 '18 at 17:36






                        $begingroup$
                        Q1: that you can tell the floats (or the orchestras) apart; or to put it differently: would swapping two floats (or two orchestras) count as a different line-up, or not? If so, your work is fine. If not - which is more likely in this context - see my answer (and the other ones!).
                        $endgroup$
                        – StackTD
                        Dec 14 '18 at 17:36













                        0












                        $begingroup$

                        1) is correct ... assuming one distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other ... personally, I couldn't take two orchestras in a row ...), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats.



                        For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          1) is correct ... assuming one distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other ... personally, I couldn't take two orchestras in a row ...), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats.



                          For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            1) is correct ... assuming one distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other ... personally, I couldn't take two orchestras in a row ...), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats.



                            For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?






                            share|cite|improve this answer









                            $endgroup$



                            1) is correct ... assuming one distinguishes between the floats (and same for the orchestras) .... for example, I can see how someone would be merely interested in the distribution of the orchestras among the floats (e.g. they want to make sure there are no two orchestras next to each other ... personally, I couldn't take two orchestras in a row ...), and from that perspective they might have no interested in distinguishing between the different orchestras, or between the different floats.



                            For 2) .. couldn't someone win two dogs? Or all? How is this raffle organized?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 14 '18 at 16:57









                            Bram28Bram28

                            62.1k44793




                            62.1k44793






























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