Number of ways of forming 10 student committee from 5 classes of 30 students each

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There are 5 classes with 30 students each. How many ways can a committee of 10 students be formed if each class has to have at least one student on the committee?



I figured that we first have to choose 5 people from each class, so there are $10^5$ options. There remain total of 29*5=145 students to choose from, and we can fill remaining 5 spots any way we want, i.e., $binom1455$ options. Thus, I thought answer would be $$10^5 dot binom1455 $$.



Apparently this is the wrong answer. I can see how the inclusion/exclusion principle also leads to the right answer. That is,
$$ 150 choose 10- 5120 choose 10 + 5 choose 290 choose 10 - 5 choose 360 choose 10 + 5 choose 430 choose 10 $$



I don't see though what I overlooked and why my answer is different?



Any advice on when to simply use inclusion/exclusion and when to use other methods?




combinatorics discrete-mathematics combinations inclusion-exclusion




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  • Shouldn't it be $9 , 8 , 7 , 6 $ in the later cases in place of $10$ ?
    – Entrepreneur
    Jul 17 at 7:39















up vote
2
down vote

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There are 5 classes with 30 students each. How many ways can a committee of 10 students be formed if each class has to have at least one student on the committee?



I figured that we first have to choose 5 people from each class, so there are $10^5$ options. There remain total of 29*5=145 students to choose from, and we can fill remaining 5 spots any way we want, i.e., $binom1455$ options. Thus, I thought answer would be $$10^5 dot binom1455 $$.



Apparently this is the wrong answer. I can see how the inclusion/exclusion principle also leads to the right answer. That is,
$$ 150 choose 10- 5120 choose 10 + 5 choose 290 choose 10 - 5 choose 360 choose 10 + 5 choose 430 choose 10 $$



I don't see though what I overlooked and why my answer is different?



Any advice on when to simply use inclusion/exclusion and when to use other methods?




combinatorics discrete-mathematics combinations inclusion-exclusion




share|cite|improve this question



















  • Shouldn't it be $9 , 8 , 7 , 6 $ in the later cases in place of $10$ ?
    – Entrepreneur
    Jul 17 at 7:39













up vote
2
down vote

favorite









up vote
2
down vote

favorite











There are 5 classes with 30 students each. How many ways can a committee of 10 students be formed if each class has to have at least one student on the committee?



I figured that we first have to choose 5 people from each class, so there are $10^5$ options. There remain total of 29*5=145 students to choose from, and we can fill remaining 5 spots any way we want, i.e., $binom1455$ options. Thus, I thought answer would be $$10^5 dot binom1455 $$.



Apparently this is the wrong answer. I can see how the inclusion/exclusion principle also leads to the right answer. That is,
$$ 150 choose 10- 5120 choose 10 + 5 choose 290 choose 10 - 5 choose 360 choose 10 + 5 choose 430 choose 10 $$



I don't see though what I overlooked and why my answer is different?



Any advice on when to simply use inclusion/exclusion and when to use other methods?




combinatorics discrete-mathematics combinations inclusion-exclusion




share|cite|improve this question











There are 5 classes with 30 students each. How many ways can a committee of 10 students be formed if each class has to have at least one student on the committee?



I figured that we first have to choose 5 people from each class, so there are $10^5$ options. There remain total of 29*5=145 students to choose from, and we can fill remaining 5 spots any way we want, i.e., $binom1455$ options. Thus, I thought answer would be $$10^5 dot binom1455 $$.



Apparently this is the wrong answer. I can see how the inclusion/exclusion principle also leads to the right answer. That is,
$$ 150 choose 10- 5120 choose 10 + 5 choose 290 choose 10 - 5 choose 360 choose 10 + 5 choose 430 choose 10 $$



I don't see though what I overlooked and why my answer is different?



Any advice on when to simply use inclusion/exclusion and when to use other methods?





combinatorics discrete-mathematics combinations inclusion-exclusion





share|cite|improve this question










share|cite|improve this question




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asked Jul 15 at 6:49









Adam G

363




363











  • Shouldn't it be $9 , 8 , 7 , 6 $ in the later cases in place of $10$ ?
    – Entrepreneur
    Jul 17 at 7:39

















  • Shouldn't it be $9 , 8 , 7 , 6 $ in the later cases in place of $10$ ?
    – Entrepreneur
    Jul 17 at 7:39
















Shouldn't it be $9 , 8 , 7 , 6 $ in the later cases in place of $10$ ?
– Entrepreneur
Jul 17 at 7:39





Shouldn't it be $9 , 8 , 7 , 6 $ in the later cases in place of $10$ ?
– Entrepreneur
Jul 17 at 7:39











2 Answers
2






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up vote
2
down vote













You are over counting.



Suppose, for a simpler example, you had two classes of three students $a,b,c,d,e,f$ and have count ways to form distinct committees of three that includes at least one student from each class.



Your method would have you count ways to pick one from $a,b,c$, one from $d,e,f$, and one from the remainder. $binom 31binom 31binom 41$ ways; $36$.   So you could pick $a$ and $d$ then $b$ and form committee $a,b,d$.   Another ways is to pick $b$ and $d$, then $a$, to form.... the same committee, $a,b,d$.



So rather, we count the total ways to select three from six and use PIE.   That is $binom 63-2binom 33+binom 03$, or $18$.



Which should equal ways to select two from $a,b,c$ and one from $d,e,f$ and ways to select one from $a,b,c$ and two from $d,e,f$.   $binom 32binom 31times 2$.   Which is indeed $18$.






share|cite|improve this answer




























    up vote
    0
    down vote













    Here is how it works without inclusion-exclusion.



    There are 7 ways of partitioning 10 into 5 positive parts. The number may be checked here https://oeis.org/A008284



    The problem is similar to a poker game hands problem but here we have 5 colors and 30 values.



    For each kind of "hand" we have to do the specific calculus.



    $case 10=6+1+1+1+1$ :



    there are $binom51$ choices of the long color. For each one, there are $binom306$ choices of the values. For each other short color there are $binom301$ choices, for a total of :



    $binom51 binom306binom301^4$



    $case 10=5+2+1+1+1$ :



    there are $binom51$ choices of the long color. There are $binom41$ choices for the color of a pair. For the long color, there are $binom305$ choices of the values. For the pair there are $binom302$ choices; the total is for a total of:



    $binom51 binom41 binom305 binom302 binom301^3$



    Let's write for the sake of comparing the complete formula:



    $$binom51 binom306binom301^4 + binom51 binom41 binom305 binom302 binom301^3 + binom 51 binom41binom304binom 303binom301^3+ binom51 binom42 binom 304 binom 302^2 binom301^2 + binom51 binom42 binom 302 binom 303^2 binom301^2 + binom51 binom41 binom 303 binom 301 binom302^3 + binom302^5 $$



    After computing all seven cases, we get the required 645666069796875. I think, given the similarity with poker where there are involved a lot of money, someone would have invented in the last centuries a shorter way of computing hands.



    The inclusion-exclusion is possible since we can manipulate that "at least one" and the calculus is shorter since the classes are equal.






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      2 Answers
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      active

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      2 Answers
      2






      active

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      active

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      up vote
      2
      down vote













      You are over counting.



      Suppose, for a simpler example, you had two classes of three students $a,b,c,d,e,f$ and have count ways to form distinct committees of three that includes at least one student from each class.



      Your method would have you count ways to pick one from $a,b,c$, one from $d,e,f$, and one from the remainder. $binom 31binom 31binom 41$ ways; $36$.   So you could pick $a$ and $d$ then $b$ and form committee $a,b,d$.   Another ways is to pick $b$ and $d$, then $a$, to form.... the same committee, $a,b,d$.



      So rather, we count the total ways to select three from six and use PIE.   That is $binom 63-2binom 33+binom 03$, or $18$.



      Which should equal ways to select two from $a,b,c$ and one from $d,e,f$ and ways to select one from $a,b,c$ and two from $d,e,f$.   $binom 32binom 31times 2$.   Which is indeed $18$.






      share|cite|improve this answer

























        up vote
        2
        down vote













        You are over counting.



        Suppose, for a simpler example, you had two classes of three students $a,b,c,d,e,f$ and have count ways to form distinct committees of three that includes at least one student from each class.



        Your method would have you count ways to pick one from $a,b,c$, one from $d,e,f$, and one from the remainder. $binom 31binom 31binom 41$ ways; $36$.   So you could pick $a$ and $d$ then $b$ and form committee $a,b,d$.   Another ways is to pick $b$ and $d$, then $a$, to form.... the same committee, $a,b,d$.



        So rather, we count the total ways to select three from six and use PIE.   That is $binom 63-2binom 33+binom 03$, or $18$.



        Which should equal ways to select two from $a,b,c$ and one from $d,e,f$ and ways to select one from $a,b,c$ and two from $d,e,f$.   $binom 32binom 31times 2$.   Which is indeed $18$.






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          You are over counting.



          Suppose, for a simpler example, you had two classes of three students $a,b,c,d,e,f$ and have count ways to form distinct committees of three that includes at least one student from each class.



          Your method would have you count ways to pick one from $a,b,c$, one from $d,e,f$, and one from the remainder. $binom 31binom 31binom 41$ ways; $36$.   So you could pick $a$ and $d$ then $b$ and form committee $a,b,d$.   Another ways is to pick $b$ and $d$, then $a$, to form.... the same committee, $a,b,d$.



          So rather, we count the total ways to select three from six and use PIE.   That is $binom 63-2binom 33+binom 03$, or $18$.



          Which should equal ways to select two from $a,b,c$ and one from $d,e,f$ and ways to select one from $a,b,c$ and two from $d,e,f$.   $binom 32binom 31times 2$.   Which is indeed $18$.






          share|cite|improve this answer













          You are over counting.



          Suppose, for a simpler example, you had two classes of three students $a,b,c,d,e,f$ and have count ways to form distinct committees of three that includes at least one student from each class.



          Your method would have you count ways to pick one from $a,b,c$, one from $d,e,f$, and one from the remainder. $binom 31binom 31binom 41$ ways; $36$.   So you could pick $a$ and $d$ then $b$ and form committee $a,b,d$.   Another ways is to pick $b$ and $d$, then $a$, to form.... the same committee, $a,b,d$.



          So rather, we count the total ways to select three from six and use PIE.   That is $binom 63-2binom 33+binom 03$, or $18$.



          Which should equal ways to select two from $a,b,c$ and one from $d,e,f$ and ways to select one from $a,b,c$ and two from $d,e,f$.   $binom 32binom 31times 2$.   Which is indeed $18$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 15 at 7:06









          Graham Kemp

          80.1k43275




          80.1k43275




















              up vote
              0
              down vote













              Here is how it works without inclusion-exclusion.



              There are 7 ways of partitioning 10 into 5 positive parts. The number may be checked here https://oeis.org/A008284



              The problem is similar to a poker game hands problem but here we have 5 colors and 30 values.



              For each kind of "hand" we have to do the specific calculus.



              $case 10=6+1+1+1+1$ :



              there are $binom51$ choices of the long color. For each one, there are $binom306$ choices of the values. For each other short color there are $binom301$ choices, for a total of :



              $binom51 binom306binom301^4$



              $case 10=5+2+1+1+1$ :



              there are $binom51$ choices of the long color. There are $binom41$ choices for the color of a pair. For the long color, there are $binom305$ choices of the values. For the pair there are $binom302$ choices; the total is for a total of:



              $binom51 binom41 binom305 binom302 binom301^3$



              Let's write for the sake of comparing the complete formula:



              $$binom51 binom306binom301^4 + binom51 binom41 binom305 binom302 binom301^3 + binom 51 binom41binom304binom 303binom301^3+ binom51 binom42 binom 304 binom 302^2 binom301^2 + binom51 binom42 binom 302 binom 303^2 binom301^2 + binom51 binom41 binom 303 binom 301 binom302^3 + binom302^5 $$



              After computing all seven cases, we get the required 645666069796875. I think, given the similarity with poker where there are involved a lot of money, someone would have invented in the last centuries a shorter way of computing hands.



              The inclusion-exclusion is possible since we can manipulate that "at least one" and the calculus is shorter since the classes are equal.






              share|cite|improve this answer



























                up vote
                0
                down vote













                Here is how it works without inclusion-exclusion.



                There are 7 ways of partitioning 10 into 5 positive parts. The number may be checked here https://oeis.org/A008284



                The problem is similar to a poker game hands problem but here we have 5 colors and 30 values.



                For each kind of "hand" we have to do the specific calculus.



                $case 10=6+1+1+1+1$ :



                there are $binom51$ choices of the long color. For each one, there are $binom306$ choices of the values. For each other short color there are $binom301$ choices, for a total of :



                $binom51 binom306binom301^4$



                $case 10=5+2+1+1+1$ :



                there are $binom51$ choices of the long color. There are $binom41$ choices for the color of a pair. For the long color, there are $binom305$ choices of the values. For the pair there are $binom302$ choices; the total is for a total of:



                $binom51 binom41 binom305 binom302 binom301^3$



                Let's write for the sake of comparing the complete formula:



                $$binom51 binom306binom301^4 + binom51 binom41 binom305 binom302 binom301^3 + binom 51 binom41binom304binom 303binom301^3+ binom51 binom42 binom 304 binom 302^2 binom301^2 + binom51 binom42 binom 302 binom 303^2 binom301^2 + binom51 binom41 binom 303 binom 301 binom302^3 + binom302^5 $$



                After computing all seven cases, we get the required 645666069796875. I think, given the similarity with poker where there are involved a lot of money, someone would have invented in the last centuries a shorter way of computing hands.



                The inclusion-exclusion is possible since we can manipulate that "at least one" and the calculus is shorter since the classes are equal.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Here is how it works without inclusion-exclusion.



                  There are 7 ways of partitioning 10 into 5 positive parts. The number may be checked here https://oeis.org/A008284



                  The problem is similar to a poker game hands problem but here we have 5 colors and 30 values.



                  For each kind of "hand" we have to do the specific calculus.



                  $case 10=6+1+1+1+1$ :



                  there are $binom51$ choices of the long color. For each one, there are $binom306$ choices of the values. For each other short color there are $binom301$ choices, for a total of :



                  $binom51 binom306binom301^4$



                  $case 10=5+2+1+1+1$ :



                  there are $binom51$ choices of the long color. There are $binom41$ choices for the color of a pair. For the long color, there are $binom305$ choices of the values. For the pair there are $binom302$ choices; the total is for a total of:



                  $binom51 binom41 binom305 binom302 binom301^3$



                  Let's write for the sake of comparing the complete formula:



                  $$binom51 binom306binom301^4 + binom51 binom41 binom305 binom302 binom301^3 + binom 51 binom41binom304binom 303binom301^3+ binom51 binom42 binom 304 binom 302^2 binom301^2 + binom51 binom42 binom 302 binom 303^2 binom301^2 + binom51 binom41 binom 303 binom 301 binom302^3 + binom302^5 $$



                  After computing all seven cases, we get the required 645666069796875. I think, given the similarity with poker where there are involved a lot of money, someone would have invented in the last centuries a shorter way of computing hands.



                  The inclusion-exclusion is possible since we can manipulate that "at least one" and the calculus is shorter since the classes are equal.






                  share|cite|improve this answer















                  Here is how it works without inclusion-exclusion.



                  There are 7 ways of partitioning 10 into 5 positive parts. The number may be checked here https://oeis.org/A008284



                  The problem is similar to a poker game hands problem but here we have 5 colors and 30 values.



                  For each kind of "hand" we have to do the specific calculus.



                  $case 10=6+1+1+1+1$ :



                  there are $binom51$ choices of the long color. For each one, there are $binom306$ choices of the values. For each other short color there are $binom301$ choices, for a total of :



                  $binom51 binom306binom301^4$



                  $case 10=5+2+1+1+1$ :



                  there are $binom51$ choices of the long color. There are $binom41$ choices for the color of a pair. For the long color, there are $binom305$ choices of the values. For the pair there are $binom302$ choices; the total is for a total of:



                  $binom51 binom41 binom305 binom302 binom301^3$



                  Let's write for the sake of comparing the complete formula:



                  $$binom51 binom306binom301^4 + binom51 binom41 binom305 binom302 binom301^3 + binom 51 binom41binom304binom 303binom301^3+ binom51 binom42 binom 304 binom 302^2 binom301^2 + binom51 binom42 binom 302 binom 303^2 binom301^2 + binom51 binom41 binom 303 binom 301 binom302^3 + binom302^5 $$



                  After computing all seven cases, we get the required 645666069796875. I think, given the similarity with poker where there are involved a lot of money, someone would have invented in the last centuries a shorter way of computing hands.



                  The inclusion-exclusion is possible since we can manipulate that "at least one" and the calculus is shorter since the classes are equal.







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 16 at 10:44


























                  answered Jul 16 at 10:35









                  nbaxter

                  45710




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