When do i use the power ^ sign in a combination question?












1












$begingroup$


My problem: If there are 5 different candies in a jar and a child wants to take out one or more candies, how many ways can this be done?



I said it is $^5C_1 -; ^5C_0 = 5-1 = 4$ ways. The $-1$ for the unwanted case using this trick:



At least/At most = total number of combinations - unwanted cases



But according to my answer sheet, it said $2^5 -1$ is the answer.



So my question is that in what situations should I use exponents and what impact does it have?






combinatorics discrete-mathematics combinations







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$endgroup$












  • $begingroup$
    The child can take 1 OR 2 OR 3 OR 4 OR 5, each with their own combinations
    $endgroup$
    – Paul
    Dec 10 '18 at 11:53
















1












$begingroup$


My problem: If there are 5 different candies in a jar and a child wants to take out one or more candies, how many ways can this be done?



I said it is $^5C_1 -; ^5C_0 = 5-1 = 4$ ways. The $-1$ for the unwanted case using this trick:



At least/At most = total number of combinations - unwanted cases



But according to my answer sheet, it said $2^5 -1$ is the answer.



So my question is that in what situations should I use exponents and what impact does it have?






combinatorics discrete-mathematics combinations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The child can take 1 OR 2 OR 3 OR 4 OR 5, each with their own combinations
    $endgroup$
    – Paul
    Dec 10 '18 at 11:53














1












1








1


1



$begingroup$


My problem: If there are 5 different candies in a jar and a child wants to take out one or more candies, how many ways can this be done?



I said it is $^5C_1 -; ^5C_0 = 5-1 = 4$ ways. The $-1$ for the unwanted case using this trick:



At least/At most = total number of combinations - unwanted cases



But according to my answer sheet, it said $2^5 -1$ is the answer.



So my question is that in what situations should I use exponents and what impact does it have?






combinatorics discrete-mathematics combinations







share|cite|improve this question











$endgroup$




My problem: If there are 5 different candies in a jar and a child wants to take out one or more candies, how many ways can this be done?



I said it is $^5C_1 -; ^5C_0 = 5-1 = 4$ ways. The $-1$ for the unwanted case using this trick:



At least/At most = total number of combinations - unwanted cases



But according to my answer sheet, it said $2^5 -1$ is the answer.



So my question is that in what situations should I use exponents and what impact does it have?






combinatorics discrete-mathematics combinations




combinatorics discrete-mathematics combinations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 11:58









amWhy

1




1










asked Dec 10 '18 at 11:49









Fred WeasleyFred Weasley

14510




14510












  • $begingroup$
    The child can take 1 OR 2 OR 3 OR 4 OR 5, each with their own combinations
    $endgroup$
    – Paul
    Dec 10 '18 at 11:53


















  • $begingroup$
    The child can take 1 OR 2 OR 3 OR 4 OR 5, each with their own combinations
    $endgroup$
    – Paul
    Dec 10 '18 at 11:53
















$begingroup$
The child can take 1 OR 2 OR 3 OR 4 OR 5, each with their own combinations
$endgroup$
– Paul
Dec 10 '18 at 11:53




$begingroup$
The child can take 1 OR 2 OR 3 OR 4 OR 5, each with their own combinations
$endgroup$
– Paul
Dec 10 '18 at 11:53










3 Answers
3






active

oldest

votes


















2












$begingroup$

Generally you use a power whenever you have several independent choices to make, and each choice has the same options.



Here, for each candy in the jar you can choose to take it, or not ($2$ options). All these decisions are independent, and there are $5$ of them, so there are $2^5$ possibilities for which candies you can take from the jar. However, this includes the case of taking none of them, so you need to subtract $1$ to get the answer you're looking for.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    First of all thank you for replying!!! Although, i am not sure what you meant by independent choices and what do mean by 'each choice has the same options'? I thought option = choice?
    $endgroup$
    – Fred Weasley
    Dec 10 '18 at 12:01










  • $begingroup$
    I meant "choice" as in "decision" - you have to decide, for each candy, whether you want it or not. So that's 5 decisions with 2 options each, and they are independent in that what you choose for candy A doesn't affect what you can do with candy B.
    $endgroup$
    – Especially Lime
    Dec 10 '18 at 12:03



















1












$begingroup$

Adding to the other answers:



$^5C_1-, ^5C_0$ doesn't make sense for the following reasons:




  • The problem states that the child takes one or more candies. None of these quantities in the expression includes data about the case where more than one candy is taken.


  • $^5C_1$ gives the number of ways to select one candy. On the other hand, $^5C_0$ is the number of ways to select $0$ candies. It doesn't make sense to subtract these two because they correspond to different events.







share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Think of it like this: The child can either take a specific candy or not take it. This means we have $2$ possibilities for whether this candy is taken or not. Given we have $5$ candies, we have $2cdot2cdot2cdot2cdot2=2^5$ ways of taking $5$ candies. Since the condition that we have $0$ candies can be ignored, we have $2^5-1$ candies. Using $^5C_1-^5C_0$ would work if we could pick only one candy. (although the $^5C_0$ is unnecessary)






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      It would be better to say that having $0$ candies is prohibited.
      $endgroup$
      – N. F. Taussig
      Dec 11 '18 at 10:21











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Generally you use a power whenever you have several independent choices to make, and each choice has the same options.



    Here, for each candy in the jar you can choose to take it, or not ($2$ options). All these decisions are independent, and there are $5$ of them, so there are $2^5$ possibilities for which candies you can take from the jar. However, this includes the case of taking none of them, so you need to subtract $1$ to get the answer you're looking for.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      First of all thank you for replying!!! Although, i am not sure what you meant by independent choices and what do mean by 'each choice has the same options'? I thought option = choice?
      $endgroup$
      – Fred Weasley
      Dec 10 '18 at 12:01










    • $begingroup$
      I meant "choice" as in "decision" - you have to decide, for each candy, whether you want it or not. So that's 5 decisions with 2 options each, and they are independent in that what you choose for candy A doesn't affect what you can do with candy B.
      $endgroup$
      – Especially Lime
      Dec 10 '18 at 12:03
















    2












    $begingroup$

    Generally you use a power whenever you have several independent choices to make, and each choice has the same options.



    Here, for each candy in the jar you can choose to take it, or not ($2$ options). All these decisions are independent, and there are $5$ of them, so there are $2^5$ possibilities for which candies you can take from the jar. However, this includes the case of taking none of them, so you need to subtract $1$ to get the answer you're looking for.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      First of all thank you for replying!!! Although, i am not sure what you meant by independent choices and what do mean by 'each choice has the same options'? I thought option = choice?
      $endgroup$
      – Fred Weasley
      Dec 10 '18 at 12:01










    • $begingroup$
      I meant "choice" as in "decision" - you have to decide, for each candy, whether you want it or not. So that's 5 decisions with 2 options each, and they are independent in that what you choose for candy A doesn't affect what you can do with candy B.
      $endgroup$
      – Especially Lime
      Dec 10 '18 at 12:03














    2












    2








    2





    $begingroup$

    Generally you use a power whenever you have several independent choices to make, and each choice has the same options.



    Here, for each candy in the jar you can choose to take it, or not ($2$ options). All these decisions are independent, and there are $5$ of them, so there are $2^5$ possibilities for which candies you can take from the jar. However, this includes the case of taking none of them, so you need to subtract $1$ to get the answer you're looking for.






    share|cite|improve this answer









    $endgroup$



    Generally you use a power whenever you have several independent choices to make, and each choice has the same options.



    Here, for each candy in the jar you can choose to take it, or not ($2$ options). All these decisions are independent, and there are $5$ of them, so there are $2^5$ possibilities for which candies you can take from the jar. However, this includes the case of taking none of them, so you need to subtract $1$ to get the answer you're looking for.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 11:56









    Especially LimeEspecially Lime

    21.9k22858




    21.9k22858












    • $begingroup$
      First of all thank you for replying!!! Although, i am not sure what you meant by independent choices and what do mean by 'each choice has the same options'? I thought option = choice?
      $endgroup$
      – Fred Weasley
      Dec 10 '18 at 12:01










    • $begingroup$
      I meant "choice" as in "decision" - you have to decide, for each candy, whether you want it or not. So that's 5 decisions with 2 options each, and they are independent in that what you choose for candy A doesn't affect what you can do with candy B.
      $endgroup$
      – Especially Lime
      Dec 10 '18 at 12:03


















    • $begingroup$
      First of all thank you for replying!!! Although, i am not sure what you meant by independent choices and what do mean by 'each choice has the same options'? I thought option = choice?
      $endgroup$
      – Fred Weasley
      Dec 10 '18 at 12:01










    • $begingroup$
      I meant "choice" as in "decision" - you have to decide, for each candy, whether you want it or not. So that's 5 decisions with 2 options each, and they are independent in that what you choose for candy A doesn't affect what you can do with candy B.
      $endgroup$
      – Especially Lime
      Dec 10 '18 at 12:03
















    $begingroup$
    First of all thank you for replying!!! Although, i am not sure what you meant by independent choices and what do mean by 'each choice has the same options'? I thought option = choice?
    $endgroup$
    – Fred Weasley
    Dec 10 '18 at 12:01




    $begingroup$
    First of all thank you for replying!!! Although, i am not sure what you meant by independent choices and what do mean by 'each choice has the same options'? I thought option = choice?
    $endgroup$
    – Fred Weasley
    Dec 10 '18 at 12:01












    $begingroup$
    I meant "choice" as in "decision" - you have to decide, for each candy, whether you want it or not. So that's 5 decisions with 2 options each, and they are independent in that what you choose for candy A doesn't affect what you can do with candy B.
    $endgroup$
    – Especially Lime
    Dec 10 '18 at 12:03




    $begingroup$
    I meant "choice" as in "decision" - you have to decide, for each candy, whether you want it or not. So that's 5 decisions with 2 options each, and they are independent in that what you choose for candy A doesn't affect what you can do with candy B.
    $endgroup$
    – Especially Lime
    Dec 10 '18 at 12:03











    1












    $begingroup$

    Adding to the other answers:



    $^5C_1-, ^5C_0$ doesn't make sense for the following reasons:




    • The problem states that the child takes one or more candies. None of these quantities in the expression includes data about the case where more than one candy is taken.


    • $^5C_1$ gives the number of ways to select one candy. On the other hand, $^5C_0$ is the number of ways to select $0$ candies. It doesn't make sense to subtract these two because they correspond to different events.







    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Adding to the other answers:



      $^5C_1-, ^5C_0$ doesn't make sense for the following reasons:




      • The problem states that the child takes one or more candies. None of these quantities in the expression includes data about the case where more than one candy is taken.


      • $^5C_1$ gives the number of ways to select one candy. On the other hand, $^5C_0$ is the number of ways to select $0$ candies. It doesn't make sense to subtract these two because they correspond to different events.







      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Adding to the other answers:



        $^5C_1-, ^5C_0$ doesn't make sense for the following reasons:




        • The problem states that the child takes one or more candies. None of these quantities in the expression includes data about the case where more than one candy is taken.


        • $^5C_1$ gives the number of ways to select one candy. On the other hand, $^5C_0$ is the number of ways to select $0$ candies. It doesn't make sense to subtract these two because they correspond to different events.







        share|cite|improve this answer











        $endgroup$



        Adding to the other answers:



        $^5C_1-, ^5C_0$ doesn't make sense for the following reasons:




        • The problem states that the child takes one or more candies. None of these quantities in the expression includes data about the case where more than one candy is taken.


        • $^5C_1$ gives the number of ways to select one candy. On the other hand, $^5C_0$ is the number of ways to select $0$ candies. It doesn't make sense to subtract these two because they correspond to different events.








        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '18 at 13:07









        amWhy

        1




        1










        answered Dec 10 '18 at 11:59









        Michael BurrMichael Burr

        26.7k23262




        26.7k23262























            0












            $begingroup$

            Think of it like this: The child can either take a specific candy or not take it. This means we have $2$ possibilities for whether this candy is taken or not. Given we have $5$ candies, we have $2cdot2cdot2cdot2cdot2=2^5$ ways of taking $5$ candies. Since the condition that we have $0$ candies can be ignored, we have $2^5-1$ candies. Using $^5C_1-^5C_0$ would work if we could pick only one candy. (although the $^5C_0$ is unnecessary)






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              It would be better to say that having $0$ candies is prohibited.
              $endgroup$
              – N. F. Taussig
              Dec 11 '18 at 10:21
















            0












            $begingroup$

            Think of it like this: The child can either take a specific candy or not take it. This means we have $2$ possibilities for whether this candy is taken or not. Given we have $5$ candies, we have $2cdot2cdot2cdot2cdot2=2^5$ ways of taking $5$ candies. Since the condition that we have $0$ candies can be ignored, we have $2^5-1$ candies. Using $^5C_1-^5C_0$ would work if we could pick only one candy. (although the $^5C_0$ is unnecessary)






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              It would be better to say that having $0$ candies is prohibited.
              $endgroup$
              – N. F. Taussig
              Dec 11 '18 at 10:21














            0












            0








            0





            $begingroup$

            Think of it like this: The child can either take a specific candy or not take it. This means we have $2$ possibilities for whether this candy is taken or not. Given we have $5$ candies, we have $2cdot2cdot2cdot2cdot2=2^5$ ways of taking $5$ candies. Since the condition that we have $0$ candies can be ignored, we have $2^5-1$ candies. Using $^5C_1-^5C_0$ would work if we could pick only one candy. (although the $^5C_0$ is unnecessary)






            share|cite|improve this answer









            $endgroup$



            Think of it like this: The child can either take a specific candy or not take it. This means we have $2$ possibilities for whether this candy is taken or not. Given we have $5$ candies, we have $2cdot2cdot2cdot2cdot2=2^5$ ways of taking $5$ candies. Since the condition that we have $0$ candies can be ignored, we have $2^5-1$ candies. Using $^5C_1-^5C_0$ would work if we could pick only one candy. (although the $^5C_0$ is unnecessary)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 12:02









            KykyKyky

            450213




            450213








            • 1




              $begingroup$
              It would be better to say that having $0$ candies is prohibited.
              $endgroup$
              – N. F. Taussig
              Dec 11 '18 at 10:21














            • 1




              $begingroup$
              It would be better to say that having $0$ candies is prohibited.
              $endgroup$
              – N. F. Taussig
              Dec 11 '18 at 10:21








            1




            1




            $begingroup$
            It would be better to say that having $0$ candies is prohibited.
            $endgroup$
            – N. F. Taussig
            Dec 11 '18 at 10:21




            $begingroup$
            It would be better to say that having $0$ candies is prohibited.
            $endgroup$
            – N. F. Taussig
            Dec 11 '18 at 10:21


















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