Plain simple combinatorics formula vs. using polynomials for counting the number of balls which can be picked...
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I am really confused by the usage of polynomials for counting problem, for instance :
Number of ways of picking $5$ balls from $5$ red, $5$ green, $5$ blue balls.
Condition : At least one of each colour has to be picked.
My approach for the above problem was quite simple, i.e. given the fact that at least one ball for each colour should be picked, we are left with $2$ more balls to be picked of any colour, so that would be: $${15-3choose 2} = {12choose 2} = 66$$
But, the source from where I am learning this course on Combinatorics had a different approach, here's a screenshot from the video where the instructor solved the given problem with the help of Polynomials, i.e. getting the coefficient of $x^5$ in the expansion of the given polynomial:
Clearly the solution I ended up with and the solution by the instructor are completely different. I have understood the solution of the instructor, but what I am still confused with is that I am not unable to understand where I want wrong with my approach?
Here is the link to the video
combinatorics polynomials combinations generating-functions
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
$endgroup$
add a comment |
$begingroup$
I am really confused by the usage of polynomials for counting problem, for instance :
Number of ways of picking $5$ balls from $5$ red, $5$ green, $5$ blue balls.
Condition : At least one of each colour has to be picked.
My approach for the above problem was quite simple, i.e. given the fact that at least one ball for each colour should be picked, we are left with $2$ more balls to be picked of any colour, so that would be: $${15-3choose 2} = {12choose 2} = 66$$
But, the source from where I am learning this course on Combinatorics had a different approach, here's a screenshot from the video where the instructor solved the given problem with the help of Polynomials, i.e. getting the coefficient of $x^5$ in the expansion of the given polynomial:
Clearly the solution I ended up with and the solution by the instructor are completely different. I have understood the solution of the instructor, but what I am still confused with is that I am not unable to understand where I want wrong with my approach?
Here is the link to the video
combinatorics polynomials combinations generating-functions
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
$endgroup$
2
$begingroup$
You cannot use $binom{12}2$, because the balls are not distinct. All that matters is how many of each color you picked, not which particular ball you picked.
$endgroup$
– Mike Earnest
Dec 3 '18 at 15:11
1
$begingroup$
Given the answer, it's clear that the problem intends balls of the same color to be indistinguishable, although that's not always clear in the wording. The solution shown uses generating functions, but you can solve this particular problem by "stars and bars" as well. You want to know how many ways you can write 5 as an ordered sum of three positive integers (#red,#blue,#green). Imagine 5 blank balls inn a row, then place bars in 2 from the 4 gaps in between adjacent balls, which can be done in C(4,2)=6 ways. Color balls by red up to the first bar, blue between bars, etc.
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– Ned
Dec 3 '18 at 15:54
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@MikeEarnest and Ned really appreciate your help. And for anyone reading this, and would want to learn more about Stars and Bars approach, here's a video for you : youtube.com/watch?v=UTCScjoPymA
$endgroup$
– The Room
Dec 4 '18 at 3:56
add a comment |
$begingroup$
I am really confused by the usage of polynomials for counting problem, for instance :
Number of ways of picking $5$ balls from $5$ red, $5$ green, $5$ blue balls.
Condition : At least one of each colour has to be picked.
My approach for the above problem was quite simple, i.e. given the fact that at least one ball for each colour should be picked, we are left with $2$ more balls to be picked of any colour, so that would be: $${15-3choose 2} = {12choose 2} = 66$$
But, the source from where I am learning this course on Combinatorics had a different approach, here's a screenshot from the video where the instructor solved the given problem with the help of Polynomials, i.e. getting the coefficient of $x^5$ in the expansion of the given polynomial:
Clearly the solution I ended up with and the solution by the instructor are completely different. I have understood the solution of the instructor, but what I am still confused with is that I am not unable to understand where I want wrong with my approach?
Here is the link to the video
combinatorics polynomials combinations generating-functions
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
$endgroup$
I am really confused by the usage of polynomials for counting problem, for instance :
Number of ways of picking $5$ balls from $5$ red, $5$ green, $5$ blue balls.
Condition : At least one of each colour has to be picked.
My approach for the above problem was quite simple, i.e. given the fact that at least one ball for each colour should be picked, we are left with $2$ more balls to be picked of any colour, so that would be: $${15-3choose 2} = {12choose 2} = 66$$
But, the source from where I am learning this course on Combinatorics had a different approach, here's a screenshot from the video where the instructor solved the given problem with the help of Polynomials, i.e. getting the coefficient of $x^5$ in the expansion of the given polynomial:
Clearly the solution I ended up with and the solution by the instructor are completely different. I have understood the solution of the instructor, but what I am still confused with is that I am not unable to understand where I want wrong with my approach?
Here is the link to the video
combinatorics polynomials combinations generating-functions
combinatorics polynomials combinations generating-functions
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
asked Dec 3 '18 at 15:05
The RoomThe Room
15017
15017
2
$begingroup$
You cannot use $binom{12}2$, because the balls are not distinct. All that matters is how many of each color you picked, not which particular ball you picked.
$endgroup$
– Mike Earnest
Dec 3 '18 at 15:11
1
$begingroup$
Given the answer, it's clear that the problem intends balls of the same color to be indistinguishable, although that's not always clear in the wording. The solution shown uses generating functions, but you can solve this particular problem by "stars and bars" as well. You want to know how many ways you can write 5 as an ordered sum of three positive integers (#red,#blue,#green). Imagine 5 blank balls inn a row, then place bars in 2 from the 4 gaps in between adjacent balls, which can be done in C(4,2)=6 ways. Color balls by red up to the first bar, blue between bars, etc.
$endgroup$
– Ned
Dec 3 '18 at 15:54
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@MikeEarnest and Ned really appreciate your help. And for anyone reading this, and would want to learn more about Stars and Bars approach, here's a video for you : youtube.com/watch?v=UTCScjoPymA
$endgroup$
– The Room
Dec 4 '18 at 3:56
add a comment |
2
$begingroup$
You cannot use $binom{12}2$, because the balls are not distinct. All that matters is how many of each color you picked, not which particular ball you picked.
$endgroup$
– Mike Earnest
Dec 3 '18 at 15:11
1
$begingroup$
Given the answer, it's clear that the problem intends balls of the same color to be indistinguishable, although that's not always clear in the wording. The solution shown uses generating functions, but you can solve this particular problem by "stars and bars" as well. You want to know how many ways you can write 5 as an ordered sum of three positive integers (#red,#blue,#green). Imagine 5 blank balls inn a row, then place bars in 2 from the 4 gaps in between adjacent balls, which can be done in C(4,2)=6 ways. Color balls by red up to the first bar, blue between bars, etc.
$endgroup$
– Ned
Dec 3 '18 at 15:54
$begingroup$
@MikeEarnest and Ned really appreciate your help. And for anyone reading this, and would want to learn more about Stars and Bars approach, here's a video for you : youtube.com/watch?v=UTCScjoPymA
$endgroup$
– The Room
Dec 4 '18 at 3:56
2
2
$begingroup$
You cannot use $binom{12}2$, because the balls are not distinct. All that matters is how many of each color you picked, not which particular ball you picked.
$endgroup$
– Mike Earnest
Dec 3 '18 at 15:11
$begingroup$
You cannot use $binom{12}2$, because the balls are not distinct. All that matters is how many of each color you picked, not which particular ball you picked.
$endgroup$
– Mike Earnest
Dec 3 '18 at 15:11
1
1
$begingroup$
Given the answer, it's clear that the problem intends balls of the same color to be indistinguishable, although that's not always clear in the wording. The solution shown uses generating functions, but you can solve this particular problem by "stars and bars" as well. You want to know how many ways you can write 5 as an ordered sum of three positive integers (#red,#blue,#green). Imagine 5 blank balls inn a row, then place bars in 2 from the 4 gaps in between adjacent balls, which can be done in C(4,2)=6 ways. Color balls by red up to the first bar, blue between bars, etc.
$endgroup$
– Ned
Dec 3 '18 at 15:54
$begingroup$
Given the answer, it's clear that the problem intends balls of the same color to be indistinguishable, although that's not always clear in the wording. The solution shown uses generating functions, but you can solve this particular problem by "stars and bars" as well. You want to know how many ways you can write 5 as an ordered sum of three positive integers (#red,#blue,#green). Imagine 5 blank balls inn a row, then place bars in 2 from the 4 gaps in between adjacent balls, which can be done in C(4,2)=6 ways. Color balls by red up to the first bar, blue between bars, etc.
$endgroup$
– Ned
Dec 3 '18 at 15:54
$begingroup$
@MikeEarnest and Ned really appreciate your help. And for anyone reading this, and would want to learn more about Stars and Bars approach, here's a video for you : youtube.com/watch?v=UTCScjoPymA
$endgroup$
– The Room
Dec 4 '18 at 3:56
$begingroup$
@MikeEarnest and Ned really appreciate your help. And for anyone reading this, and would want to learn more about Stars and Bars approach, here's a video for you : youtube.com/watch?v=UTCScjoPymA
$endgroup$
– The Room
Dec 4 '18 at 3:56
add a comment |
1 Answer
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When dealing with counting problems you have to be very careful to use the correct construct; as was pointed out in a comment, using binomial coefficients implicitly assumes that each ball also has a number written on it to make it distinct from the others (in particular, distinct from the others of the same color), which is not what we are actually trying to count; you will find vastly over-inflated numbers by using binomial coefficients.
So what construct can you use for this problem? The best place to start is the Twelvefold way. Also in the comments is the correct interpretation of your problem (stars and bars), which you can find in the linked article (though there is difficulty with extremal cases of $n=0$).
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
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add a comment |
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$begingroup$
When dealing with counting problems you have to be very careful to use the correct construct; as was pointed out in a comment, using binomial coefficients implicitly assumes that each ball also has a number written on it to make it distinct from the others (in particular, distinct from the others of the same color), which is not what we are actually trying to count; you will find vastly over-inflated numbers by using binomial coefficients.
So what construct can you use for this problem? The best place to start is the Twelvefold way. Also in the comments is the correct interpretation of your problem (stars and bars), which you can find in the linked article (though there is difficulty with extremal cases of $n=0$).
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
$endgroup$
add a comment |
$begingroup$
When dealing with counting problems you have to be very careful to use the correct construct; as was pointed out in a comment, using binomial coefficients implicitly assumes that each ball also has a number written on it to make it distinct from the others (in particular, distinct from the others of the same color), which is not what we are actually trying to count; you will find vastly over-inflated numbers by using binomial coefficients.
So what construct can you use for this problem? The best place to start is the Twelvefold way. Also in the comments is the correct interpretation of your problem (stars and bars), which you can find in the linked article (though there is difficulty with extremal cases of $n=0$).
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
$endgroup$
add a comment |
$begingroup$
When dealing with counting problems you have to be very careful to use the correct construct; as was pointed out in a comment, using binomial coefficients implicitly assumes that each ball also has a number written on it to make it distinct from the others (in particular, distinct from the others of the same color), which is not what we are actually trying to count; you will find vastly over-inflated numbers by using binomial coefficients.
So what construct can you use for this problem? The best place to start is the Twelvefold way. Also in the comments is the correct interpretation of your problem (stars and bars), which you can find in the linked article (though there is difficulty with extremal cases of $n=0$).
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
$endgroup$
When dealing with counting problems you have to be very careful to use the correct construct; as was pointed out in a comment, using binomial coefficients implicitly assumes that each ball also has a number written on it to make it distinct from the others (in particular, distinct from the others of the same color), which is not what we are actually trying to count; you will find vastly over-inflated numbers by using binomial coefficients.
So what construct can you use for this problem? The best place to start is the Twelvefold way. Also in the comments is the correct interpretation of your problem (stars and bars), which you can find in the linked article (though there is difficulty with extremal cases of $n=0$).
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
answered Dec 3 '18 at 16:07
ImNotTheSaxManImNotTheSaxMan
1172
1172
add a comment |
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$begingroup$
You cannot use $binom{12}2$, because the balls are not distinct. All that matters is how many of each color you picked, not which particular ball you picked.
$endgroup$
– Mike Earnest
Dec 3 '18 at 15:11
1
$begingroup$
Given the answer, it's clear that the problem intends balls of the same color to be indistinguishable, although that's not always clear in the wording. The solution shown uses generating functions, but you can solve this particular problem by "stars and bars" as well. You want to know how many ways you can write 5 as an ordered sum of three positive integers (#red,#blue,#green). Imagine 5 blank balls inn a row, then place bars in 2 from the 4 gaps in between adjacent balls, which can be done in C(4,2)=6 ways. Color balls by red up to the first bar, blue between bars, etc.
$endgroup$
– Ned
Dec 3 '18 at 15:54
$begingroup$
@MikeEarnest and Ned really appreciate your help. And for anyone reading this, and would want to learn more about Stars and Bars approach, here's a video for you : youtube.com/watch?v=UTCScjoPymA
$endgroup$
– The Room
Dec 4 '18 at 3:56