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How many possible spell combinations for a wizard brawling game?



The Next CEO of Stack OverflowA formula for combinations?How can I create a subset of the 'most dissimilar' combinations from a set of possible combinations?Possible number of combinationsDifferent Possible Combinations from Different SetsMost efficient way to calculate possible combinationsTotal possible combinations of variablesHow would you work out these combinations?Mathematical formula for calculating combinations of different size between setsLooking for a counting function for combinations of optional and non optional sets(Betting) How many different full time score-combinations can it possibly be out of a 3 game-combination in football?










0












$begingroup$


I'm working on a game that involves drafting different spells and I'm trying to figure out how many combinations are possible.



Simply put, each spell has an element and a class:



  • There are $10$ elements (Fire, Water, Air, Earth, etc.)

  • There are $7$ different classes (Primary, Movement, Melee, Defensive, etc.)

  • There is one spell for each element/class combination, therefore $70$ spells

  • By the end of the game a wizard will have $7$ spells: one of each class

My questions are:



  1. I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?


  2. If the drafting was restricted to a random $4$ of the $10$ elements each game, how many possible combinations would there be?



    EDIT: You would still be drafting $7$ classes of spells, but any combinations with more than $4$ different elements would be illegal. A single game would have $4^7$ combinations, but I'm wondering about all possible games using $4$ out of $10$ elements.











share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I'm working on a game that involves drafting different spells and I'm trying to figure out how many combinations are possible.



    Simply put, each spell has an element and a class:



    • There are $10$ elements (Fire, Water, Air, Earth, etc.)

    • There are $7$ different classes (Primary, Movement, Melee, Defensive, etc.)

    • There is one spell for each element/class combination, therefore $70$ spells

    • By the end of the game a wizard will have $7$ spells: one of each class

    My questions are:



    1. I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?


    2. If the drafting was restricted to a random $4$ of the $10$ elements each game, how many possible combinations would there be?



      EDIT: You would still be drafting $7$ classes of spells, but any combinations with more than $4$ different elements would be illegal. A single game would have $4^7$ combinations, but I'm wondering about all possible games using $4$ out of $10$ elements.











    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I'm working on a game that involves drafting different spells and I'm trying to figure out how many combinations are possible.



      Simply put, each spell has an element and a class:



      • There are $10$ elements (Fire, Water, Air, Earth, etc.)

      • There are $7$ different classes (Primary, Movement, Melee, Defensive, etc.)

      • There is one spell for each element/class combination, therefore $70$ spells

      • By the end of the game a wizard will have $7$ spells: one of each class

      My questions are:



      1. I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?


      2. If the drafting was restricted to a random $4$ of the $10$ elements each game, how many possible combinations would there be?



        EDIT: You would still be drafting $7$ classes of spells, but any combinations with more than $4$ different elements would be illegal. A single game would have $4^7$ combinations, but I'm wondering about all possible games using $4$ out of $10$ elements.











      share|cite|improve this question











      $endgroup$




      I'm working on a game that involves drafting different spells and I'm trying to figure out how many combinations are possible.



      Simply put, each spell has an element and a class:



      • There are $10$ elements (Fire, Water, Air, Earth, etc.)

      • There are $7$ different classes (Primary, Movement, Melee, Defensive, etc.)

      • There is one spell for each element/class combination, therefore $70$ spells

      • By the end of the game a wizard will have $7$ spells: one of each class

      My questions are:



      1. I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?


      2. If the drafting was restricted to a random $4$ of the $10$ elements each game, how many possible combinations would there be?



        EDIT: You would still be drafting $7$ classes of spells, but any combinations with more than $4$ different elements would be illegal. A single game would have $4^7$ combinations, but I'm wondering about all possible games using $4$ out of $10$ elements.








      combinations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 19 at 21:25









      MarianD

      1,8961617




      1,8961617










      asked Mar 19 at 20:43









      Brett PenningsBrett Pennings

      1035




      1035




















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$


          I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?




          Yes.




          If the drafting was restricted to a random 4 of the 10 elements each game, how many possible combinations would there be?




          First, choose which four elements are drafted, then choose a spell of each class in the allowable elements. That's $binom104cdot 4^7$.



          But wait - we've overcounted. If we only used three of the elements in our final spell list, then there's no way of telling which other element was drafted but not used. There are seven possibilities, so those are counted seven times each, and we'll need to compensate. Subtract six times the number of ways as if we drafted three elements: $-6cdotbinom103cdot 3^7$.



          Then, this repeats for two-element spell lists. We've counted them $binom82=28$ ways from four-element drafts and $-6cdot 8=-48$ ways from three-element drafts. To compensate, we add back $21$ times the number of ways as if we drafted two elements: $21cdotbinom102cdot 2^7$.



          Finally, one-element spell lists. They're counted $binom93 = 84$ ways from four element drafts, $-6cdotbinom92=-216$ ways from three-element drafts, and $21cdot 9=189$ ways from two-element drafts. To compensate, we subtract $189-216+84-1=-56$ times the number of wats as if we drafted one element: $-56cdotbinom101cdot 1^7$.



          Overall, that's
          $$binom104cdot 4^7 -6cdotbinom103cdot 3^7 + 21cdotbinom102cdot 2^7 -56cdotbinom101cdot 1^7 = 1986400$$
          valid spell lists. Plenty of variety to work with.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Ah! $binom104cdot 4^7$ was my original guess, but that was more than the unrestricted element count so I knew it was wrong. Also do you mean "if we drafted one element:" in your last paragraph?
            $endgroup$
            – Brett Pennings
            Mar 19 at 21:34










          • $begingroup$
            Yes, that's what I mean. Corrected.
            $endgroup$
            – jmerry
            Mar 20 at 1:12











          Your Answer





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          1 Answer
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          1 Answer
          1






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          active

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          2












          $begingroup$


          I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?




          Yes.




          If the drafting was restricted to a random 4 of the 10 elements each game, how many possible combinations would there be?




          First, choose which four elements are drafted, then choose a spell of each class in the allowable elements. That's $binom104cdot 4^7$.



          But wait - we've overcounted. If we only used three of the elements in our final spell list, then there's no way of telling which other element was drafted but not used. There are seven possibilities, so those are counted seven times each, and we'll need to compensate. Subtract six times the number of ways as if we drafted three elements: $-6cdotbinom103cdot 3^7$.



          Then, this repeats for two-element spell lists. We've counted them $binom82=28$ ways from four-element drafts and $-6cdot 8=-48$ ways from three-element drafts. To compensate, we add back $21$ times the number of ways as if we drafted two elements: $21cdotbinom102cdot 2^7$.



          Finally, one-element spell lists. They're counted $binom93 = 84$ ways from four element drafts, $-6cdotbinom92=-216$ ways from three-element drafts, and $21cdot 9=189$ ways from two-element drafts. To compensate, we subtract $189-216+84-1=-56$ times the number of wats as if we drafted one element: $-56cdotbinom101cdot 1^7$.



          Overall, that's
          $$binom104cdot 4^7 -6cdotbinom103cdot 3^7 + 21cdotbinom102cdot 2^7 -56cdotbinom101cdot 1^7 = 1986400$$
          valid spell lists. Plenty of variety to work with.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Ah! $binom104cdot 4^7$ was my original guess, but that was more than the unrestricted element count so I knew it was wrong. Also do you mean "if we drafted one element:" in your last paragraph?
            $endgroup$
            – Brett Pennings
            Mar 19 at 21:34










          • $begingroup$
            Yes, that's what I mean. Corrected.
            $endgroup$
            – jmerry
            Mar 20 at 1:12















          2












          $begingroup$


          I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?




          Yes.




          If the drafting was restricted to a random 4 of the 10 elements each game, how many possible combinations would there be?




          First, choose which four elements are drafted, then choose a spell of each class in the allowable elements. That's $binom104cdot 4^7$.



          But wait - we've overcounted. If we only used three of the elements in our final spell list, then there's no way of telling which other element was drafted but not used. There are seven possibilities, so those are counted seven times each, and we'll need to compensate. Subtract six times the number of ways as if we drafted three elements: $-6cdotbinom103cdot 3^7$.



          Then, this repeats for two-element spell lists. We've counted them $binom82=28$ ways from four-element drafts and $-6cdot 8=-48$ ways from three-element drafts. To compensate, we add back $21$ times the number of ways as if we drafted two elements: $21cdotbinom102cdot 2^7$.



          Finally, one-element spell lists. They're counted $binom93 = 84$ ways from four element drafts, $-6cdotbinom92=-216$ ways from three-element drafts, and $21cdot 9=189$ ways from two-element drafts. To compensate, we subtract $189-216+84-1=-56$ times the number of wats as if we drafted one element: $-56cdotbinom101cdot 1^7$.



          Overall, that's
          $$binom104cdot 4^7 -6cdotbinom103cdot 3^7 + 21cdotbinom102cdot 2^7 -56cdotbinom101cdot 1^7 = 1986400$$
          valid spell lists. Plenty of variety to work with.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Ah! $binom104cdot 4^7$ was my original guess, but that was more than the unrestricted element count so I knew it was wrong. Also do you mean "if we drafted one element:" in your last paragraph?
            $endgroup$
            – Brett Pennings
            Mar 19 at 21:34










          • $begingroup$
            Yes, that's what I mean. Corrected.
            $endgroup$
            – jmerry
            Mar 20 at 1:12













          2












          2








          2





          $begingroup$


          I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?




          Yes.




          If the drafting was restricted to a random 4 of the 10 elements each game, how many possible combinations would there be?




          First, choose which four elements are drafted, then choose a spell of each class in the allowable elements. That's $binom104cdot 4^7$.



          But wait - we've overcounted. If we only used three of the elements in our final spell list, then there's no way of telling which other element was drafted but not used. There are seven possibilities, so those are counted seven times each, and we'll need to compensate. Subtract six times the number of ways as if we drafted three elements: $-6cdotbinom103cdot 3^7$.



          Then, this repeats for two-element spell lists. We've counted them $binom82=28$ ways from four-element drafts and $-6cdot 8=-48$ ways from three-element drafts. To compensate, we add back $21$ times the number of ways as if we drafted two elements: $21cdotbinom102cdot 2^7$.



          Finally, one-element spell lists. They're counted $binom93 = 84$ ways from four element drafts, $-6cdotbinom92=-216$ ways from three-element drafts, and $21cdot 9=189$ ways from two-element drafts. To compensate, we subtract $189-216+84-1=-56$ times the number of wats as if we drafted one element: $-56cdotbinom101cdot 1^7$.



          Overall, that's
          $$binom104cdot 4^7 -6cdotbinom103cdot 3^7 + 21cdotbinom102cdot 2^7 -56cdotbinom101cdot 1^7 = 1986400$$
          valid spell lists. Plenty of variety to work with.






          share|cite|improve this answer











          $endgroup$




          I believe there are $10^7$ combinations if there are no element restrictions. Is this correct?




          Yes.




          If the drafting was restricted to a random 4 of the 10 elements each game, how many possible combinations would there be?




          First, choose which four elements are drafted, then choose a spell of each class in the allowable elements. That's $binom104cdot 4^7$.



          But wait - we've overcounted. If we only used three of the elements in our final spell list, then there's no way of telling which other element was drafted but not used. There are seven possibilities, so those are counted seven times each, and we'll need to compensate. Subtract six times the number of ways as if we drafted three elements: $-6cdotbinom103cdot 3^7$.



          Then, this repeats for two-element spell lists. We've counted them $binom82=28$ ways from four-element drafts and $-6cdot 8=-48$ ways from three-element drafts. To compensate, we add back $21$ times the number of ways as if we drafted two elements: $21cdotbinom102cdot 2^7$.



          Finally, one-element spell lists. They're counted $binom93 = 84$ ways from four element drafts, $-6cdotbinom92=-216$ ways from three-element drafts, and $21cdot 9=189$ ways from two-element drafts. To compensate, we subtract $189-216+84-1=-56$ times the number of wats as if we drafted one element: $-56cdotbinom101cdot 1^7$.



          Overall, that's
          $$binom104cdot 4^7 -6cdotbinom103cdot 3^7 + 21cdotbinom102cdot 2^7 -56cdotbinom101cdot 1^7 = 1986400$$
          valid spell lists. Plenty of variety to work with.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 20 at 1:11

























          answered Mar 19 at 21:20









          jmerryjmerry

          16.9k11633




          16.9k11633











          • $begingroup$
            Ah! $binom104cdot 4^7$ was my original guess, but that was more than the unrestricted element count so I knew it was wrong. Also do you mean "if we drafted one element:" in your last paragraph?
            $endgroup$
            – Brett Pennings
            Mar 19 at 21:34










          • $begingroup$
            Yes, that's what I mean. Corrected.
            $endgroup$
            – jmerry
            Mar 20 at 1:12
















          • $begingroup$
            Ah! $binom104cdot 4^7$ was my original guess, but that was more than the unrestricted element count so I knew it was wrong. Also do you mean "if we drafted one element:" in your last paragraph?
            $endgroup$
            – Brett Pennings
            Mar 19 at 21:34










          • $begingroup$
            Yes, that's what I mean. Corrected.
            $endgroup$
            – jmerry
            Mar 20 at 1:12















          $begingroup$
          Ah! $binom104cdot 4^7$ was my original guess, but that was more than the unrestricted element count so I knew it was wrong. Also do you mean "if we drafted one element:" in your last paragraph?
          $endgroup$
          – Brett Pennings
          Mar 19 at 21:34




          $begingroup$
          Ah! $binom104cdot 4^7$ was my original guess, but that was more than the unrestricted element count so I knew it was wrong. Also do you mean "if we drafted one element:" in your last paragraph?
          $endgroup$
          – Brett Pennings
          Mar 19 at 21:34












          $begingroup$
          Yes, that's what I mean. Corrected.
          $endgroup$
          – jmerry
          Mar 20 at 1:12




          $begingroup$
          Yes, that's what I mean. Corrected.
          $endgroup$
          – jmerry
          Mar 20 at 1:12

















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