I have $5$ veggie items and $4$ meat items. How many different plates can I make?

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Janet must select three different items for each dinner she will serve. The items are to be chosen from among $5$ different vegetarian and $4$ different meat selections. If at least one of the selections must be vegetarian,
how many different dinners could Jane create?




The answer is $80$, I came up with a solution but I don't think it's the right way to think about the problem, is there a correct way to essentially do it?



One of the plates has to be a veggie product, so suppose to we have only have two plated with $4$ veggie options and $4$ meat options, then we have a total of $16$ possible combinations.
Thus, we introduce the $3^rd$ plate, and because this plate has to be veggie, and out of $5$ possible veggie options, we have then $16times 5 = 80$ different plates. Any other ways to think about it?






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    Janet must select three different items for each dinner she will serve. The items are to be chosen from among $5$ different vegetarian and $4$ different meat selections. If at least one of the selections must be vegetarian,
    how many different dinners could Jane create?




    The answer is $80$, I came up with a solution but I don't think it's the right way to think about the problem, is there a correct way to essentially do it?



    One of the plates has to be a veggie product, so suppose to we have only have two plated with $4$ veggie options and $4$ meat options, then we have a total of $16$ possible combinations.
    Thus, we introduce the $3^rd$ plate, and because this plate has to be veggie, and out of $5$ possible veggie options, we have then $16times 5 = 80$ different plates. Any other ways to think about it?






    combinatorics combinations







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      Janet must select three different items for each dinner she will serve. The items are to be chosen from among $5$ different vegetarian and $4$ different meat selections. If at least one of the selections must be vegetarian,
      how many different dinners could Jane create?




      The answer is $80$, I came up with a solution but I don't think it's the right way to think about the problem, is there a correct way to essentially do it?



      One of the plates has to be a veggie product, so suppose to we have only have two plated with $4$ veggie options and $4$ meat options, then we have a total of $16$ possible combinations.
      Thus, we introduce the $3^rd$ plate, and because this plate has to be veggie, and out of $5$ possible veggie options, we have then $16times 5 = 80$ different plates. Any other ways to think about it?






      combinatorics combinations







      share|cite|improve this question
















      Janet must select three different items for each dinner she will serve. The items are to be chosen from among $5$ different vegetarian and $4$ different meat selections. If at least one of the selections must be vegetarian,
      how many different dinners could Jane create?




      The answer is $80$, I came up with a solution but I don't think it's the right way to think about the problem, is there a correct way to essentially do it?



      One of the plates has to be a veggie product, so suppose to we have only have two plated with $4$ veggie options and $4$ meat options, then we have a total of $16$ possible combinations.
      Thus, we introduce the $3^rd$ plate, and because this plate has to be veggie, and out of $5$ possible veggie options, we have then $16times 5 = 80$ different plates. Any other ways to think about it?






      combinatorics combinations




      combinatorics combinations






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      edited Oct 2 '15 at 16:49







      user249332

















      asked Oct 2 '15 at 16:26









      mjo

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          It is simple. We can choose any $3$ item out of all the $5+4=9$ dishes in $colorred9choose3=colorblue84$ ways.



          Then, consider the number dishes, that there are no vegetarian dishes, i.e. only non-veg dishes. That can be done in $colorred4choose 3=colorblue4$ ways.



          Then, the number of combinations of dishes that contains at least one veg dishes is $$colorred9choose3-4choose 3=colorblue84-4=colornavyfbox80$$



          I am editing this for @ Jhon's satisfaction.



          If, from first I had counted the veg dishes, then the approach is the following.



          First, I consider that there were $1$ veg and $2$ non-veg dishes. It can happen in $$colorred5choose14choose2=colorgreen5times6=colorblue30tag 1$$ ways.



          If, there were $2$ veg and $1$ non-veg dishes, then this can happen in $$colorred5choose24choose1=colorgreen10times4=colorblue40tag2$$ ways.



          If, there were all $3$ veg dishes, then, it could happen in $$colorred5choose34choose0=colorgreen10times 1=colorblue10tag3$$ ways.



          As, you see, there are no other cases. Hence our desired result is $$colorblue30+40+10=colornavyfbox80$$






          share|cite|improve this answer






















          • How would one go if you were to count from the beginning for the veggie dishes ?Btw +1
            – Nameless
            Oct 2 '15 at 17:23










          • @Jhon, Look the edited version. I think, it will appease your query.
            – user249332
            Oct 2 '15 at 18:37










          • thank you @subhadeep Dey,unfortunately i cant give more than one +1
            – Nameless
            Oct 3 '15 at 18:17










          • You need not do that @Jhon. By the way, you are welcome.
            – user249332
            Oct 3 '15 at 18:20










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          1 Answer
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          active

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          up vote
          3
          down vote













          It is simple. We can choose any $3$ item out of all the $5+4=9$ dishes in $colorred9choose3=colorblue84$ ways.



          Then, consider the number dishes, that there are no vegetarian dishes, i.e. only non-veg dishes. That can be done in $colorred4choose 3=colorblue4$ ways.



          Then, the number of combinations of dishes that contains at least one veg dishes is $$colorred9choose3-4choose 3=colorblue84-4=colornavyfbox80$$



          I am editing this for @ Jhon's satisfaction.



          If, from first I had counted the veg dishes, then the approach is the following.



          First, I consider that there were $1$ veg and $2$ non-veg dishes. It can happen in $$colorred5choose14choose2=colorgreen5times6=colorblue30tag 1$$ ways.



          If, there were $2$ veg and $1$ non-veg dishes, then this can happen in $$colorred5choose24choose1=colorgreen10times4=colorblue40tag2$$ ways.



          If, there were all $3$ veg dishes, then, it could happen in $$colorred5choose34choose0=colorgreen10times 1=colorblue10tag3$$ ways.



          As, you see, there are no other cases. Hence our desired result is $$colorblue30+40+10=colornavyfbox80$$






          share|cite|improve this answer






















          • How would one go if you were to count from the beginning for the veggie dishes ?Btw +1
            – Nameless
            Oct 2 '15 at 17:23










          • @Jhon, Look the edited version. I think, it will appease your query.
            – user249332
            Oct 2 '15 at 18:37










          • thank you @subhadeep Dey,unfortunately i cant give more than one +1
            – Nameless
            Oct 3 '15 at 18:17










          • You need not do that @Jhon. By the way, you are welcome.
            – user249332
            Oct 3 '15 at 18:20














          up vote
          3
          down vote













          It is simple. We can choose any $3$ item out of all the $5+4=9$ dishes in $colorred9choose3=colorblue84$ ways.



          Then, consider the number dishes, that there are no vegetarian dishes, i.e. only non-veg dishes. That can be done in $colorred4choose 3=colorblue4$ ways.



          Then, the number of combinations of dishes that contains at least one veg dishes is $$colorred9choose3-4choose 3=colorblue84-4=colornavyfbox80$$



          I am editing this for @ Jhon's satisfaction.



          If, from first I had counted the veg dishes, then the approach is the following.



          First, I consider that there were $1$ veg and $2$ non-veg dishes. It can happen in $$colorred5choose14choose2=colorgreen5times6=colorblue30tag 1$$ ways.



          If, there were $2$ veg and $1$ non-veg dishes, then this can happen in $$colorred5choose24choose1=colorgreen10times4=colorblue40tag2$$ ways.



          If, there were all $3$ veg dishes, then, it could happen in $$colorred5choose34choose0=colorgreen10times 1=colorblue10tag3$$ ways.



          As, you see, there are no other cases. Hence our desired result is $$colorblue30+40+10=colornavyfbox80$$






          share|cite|improve this answer






















          • How would one go if you were to count from the beginning for the veggie dishes ?Btw +1
            – Nameless
            Oct 2 '15 at 17:23










          • @Jhon, Look the edited version. I think, it will appease your query.
            – user249332
            Oct 2 '15 at 18:37










          • thank you @subhadeep Dey,unfortunately i cant give more than one +1
            – Nameless
            Oct 3 '15 at 18:17










          • You need not do that @Jhon. By the way, you are welcome.
            – user249332
            Oct 3 '15 at 18:20












          up vote
          3
          down vote










          up vote
          3
          down vote









          It is simple. We can choose any $3$ item out of all the $5+4=9$ dishes in $colorred9choose3=colorblue84$ ways.



          Then, consider the number dishes, that there are no vegetarian dishes, i.e. only non-veg dishes. That can be done in $colorred4choose 3=colorblue4$ ways.



          Then, the number of combinations of dishes that contains at least one veg dishes is $$colorred9choose3-4choose 3=colorblue84-4=colornavyfbox80$$



          I am editing this for @ Jhon's satisfaction.



          If, from first I had counted the veg dishes, then the approach is the following.



          First, I consider that there were $1$ veg and $2$ non-veg dishes. It can happen in $$colorred5choose14choose2=colorgreen5times6=colorblue30tag 1$$ ways.



          If, there were $2$ veg and $1$ non-veg dishes, then this can happen in $$colorred5choose24choose1=colorgreen10times4=colorblue40tag2$$ ways.



          If, there were all $3$ veg dishes, then, it could happen in $$colorred5choose34choose0=colorgreen10times 1=colorblue10tag3$$ ways.



          As, you see, there are no other cases. Hence our desired result is $$colorblue30+40+10=colornavyfbox80$$






          share|cite|improve this answer














          It is simple. We can choose any $3$ item out of all the $5+4=9$ dishes in $colorred9choose3=colorblue84$ ways.



          Then, consider the number dishes, that there are no vegetarian dishes, i.e. only non-veg dishes. That can be done in $colorred4choose 3=colorblue4$ ways.



          Then, the number of combinations of dishes that contains at least one veg dishes is $$colorred9choose3-4choose 3=colorblue84-4=colornavyfbox80$$



          I am editing this for @ Jhon's satisfaction.



          If, from first I had counted the veg dishes, then the approach is the following.



          First, I consider that there were $1$ veg and $2$ non-veg dishes. It can happen in $$colorred5choose14choose2=colorgreen5times6=colorblue30tag 1$$ ways.



          If, there were $2$ veg and $1$ non-veg dishes, then this can happen in $$colorred5choose24choose1=colorgreen10times4=colorblue40tag2$$ ways.



          If, there were all $3$ veg dishes, then, it could happen in $$colorred5choose34choose0=colorgreen10times 1=colorblue10tag3$$ ways.



          As, you see, there are no other cases. Hence our desired result is $$colorblue30+40+10=colornavyfbox80$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Oct 2 '15 at 18:36

























          answered Oct 2 '15 at 16:37







          user249332


















          • How would one go if you were to count from the beginning for the veggie dishes ?Btw +1
            – Nameless
            Oct 2 '15 at 17:23










          • @Jhon, Look the edited version. I think, it will appease your query.
            – user249332
            Oct 2 '15 at 18:37










          • thank you @subhadeep Dey,unfortunately i cant give more than one +1
            – Nameless
            Oct 3 '15 at 18:17










          • You need not do that @Jhon. By the way, you are welcome.
            – user249332
            Oct 3 '15 at 18:20
















          • How would one go if you were to count from the beginning for the veggie dishes ?Btw +1
            – Nameless
            Oct 2 '15 at 17:23










          • @Jhon, Look the edited version. I think, it will appease your query.
            – user249332
            Oct 2 '15 at 18:37










          • thank you @subhadeep Dey,unfortunately i cant give more than one +1
            – Nameless
            Oct 3 '15 at 18:17










          • You need not do that @Jhon. By the way, you are welcome.
            – user249332
            Oct 3 '15 at 18:20















          How would one go if you were to count from the beginning for the veggie dishes ?Btw +1
          – Nameless
          Oct 2 '15 at 17:23




          How would one go if you were to count from the beginning for the veggie dishes ?Btw +1
          – Nameless
          Oct 2 '15 at 17:23












          @Jhon, Look the edited version. I think, it will appease your query.
          – user249332
          Oct 2 '15 at 18:37




          @Jhon, Look the edited version. I think, it will appease your query.
          – user249332
          Oct 2 '15 at 18:37












          thank you @subhadeep Dey,unfortunately i cant give more than one +1
          – Nameless
          Oct 3 '15 at 18:17




          thank you @subhadeep Dey,unfortunately i cant give more than one +1
          – Nameless
          Oct 3 '15 at 18:17












          You need not do that @Jhon. By the way, you are welcome.
          – user249332
          Oct 3 '15 at 18:20




          You need not do that @Jhon. By the way, you are welcome.
          – user249332
          Oct 3 '15 at 18:20

















           

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