How many are the non-negative integer solutions of 𝑥 + 𝑦 + 𝑤 + 𝑧 = 16 where x < y?Combinations and Permutations. Number of integer solutionsHow many integer solutions are there to the equation :|x|+|y|+|z|=15. How many integer solutions do exist?How many solutions does the equation x + y + w + z = 15 have if x, y, w, z are all non-negative integers?Combinatorics: How many non-negative integer solutions are there to each of the following equations:How many solutions does the equation $x+y+z=17$ have where $x,y,z$ are non negative integers?How many integer-valued solutions are there?How many non-negative integer solutions are there for the equation $ax + by + cz + … leq C$?How many integer solutions of $x_1+x_2+x_3+x_4=28$ are there with $-10leq x_ileq20$?How many integer solutions with negative numbers?
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How many are the non-negative integer solutions of 𝑥 + 𝑦 + 𝑤 + 𝑧 = 16 where x
Combinations and Permutations. Number of integer solutionsHow many integer solutions are there to the equation :|x|+|y|+|z|=15. How many integer solutions do exist?How many solutions does the equation x + y + w + z = 15 have if x, y, w, z are all non-negative integers?Combinatorics: How many non-negative integer solutions are there to each of the following equations:How many solutions does the equation $x+y+z=17$ have where $x,y,z$ are non negative integers?How many integer-valued solutions are there?How many non-negative integer solutions are there for the equation $ax + by + cz + … leq C$?How many integer solutions of $x_1+x_2+x_3+x_4=28$ are there with $-10leq x_ileq20$?How many integer solutions with negative numbers?
$begingroup$
Anyone can explain how to think to aproach this type of problem?
The answer is 444.
combinatorics discrete-mathematics permutations
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Anyone can explain how to think to aproach this type of problem?
The answer is 444.
combinatorics discrete-mathematics permutations
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
It's half the number of solutions for $xneq y$. If you find the number of solution for $x = y$, so the number of solutions of $2x + w + z =16$ then you just need to subtract that from the total number of solutions of $x+y+w+z$ and divide that by 2.
$endgroup$
– Count Iblis
2 hours ago
$begingroup$
Make that an answer!
$endgroup$
– Toby Mak
2 hours ago
add a comment |
$begingroup$
Anyone can explain how to think to aproach this type of problem?
The answer is 444.
combinatorics discrete-mathematics permutations
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Anyone can explain how to think to aproach this type of problem?
The answer is 444.
combinatorics discrete-mathematics permutations
combinatorics discrete-mathematics permutations
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
asked 2 hours ago
Nícolas Georgeos MantzosNícolas Georgeos Mantzos
61
61
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nícolas Georgeos Mantzos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
It's half the number of solutions for $xneq y$. If you find the number of solution for $x = y$, so the number of solutions of $2x + w + z =16$ then you just need to subtract that from the total number of solutions of $x+y+w+z$ and divide that by 2.
$endgroup$
– Count Iblis
2 hours ago
$begingroup$
Make that an answer!
$endgroup$
– Toby Mak
2 hours ago
add a comment |
3
$begingroup$
It's half the number of solutions for $xneq y$. If you find the number of solution for $x = y$, so the number of solutions of $2x + w + z =16$ then you just need to subtract that from the total number of solutions of $x+y+w+z$ and divide that by 2.
$endgroup$
– Count Iblis
2 hours ago
$begingroup$
Make that an answer!
$endgroup$
– Toby Mak
2 hours ago
3
3
$begingroup$
It's half the number of solutions for $xneq y$. If you find the number of solution for $x = y$, so the number of solutions of $2x + w + z =16$ then you just need to subtract that from the total number of solutions of $x+y+w+z$ and divide that by 2.
$endgroup$
– Count Iblis
2 hours ago
$begingroup$
It's half the number of solutions for $xneq y$. If you find the number of solution for $x = y$, so the number of solutions of $2x + w + z =16$ then you just need to subtract that from the total number of solutions of $x+y+w+z$ and divide that by 2.
$endgroup$
– Count Iblis
2 hours ago
$begingroup$
Make that an answer!
$endgroup$
– Toby Mak
2 hours ago
$begingroup$
Make that an answer!
$endgroup$
– Toby Mak
2 hours ago
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I can't comment so this answer is more of a comment. You use stars and bars. You have three bars | | | where each space represents one of x, y, w, or z. And you have 16 stars. So you count the number of distinct ways to rearrange the three bars and 16 stars, which is (3+16)!/(3! * 16!) = 969.
Next, find the number of cases where x = y (x = y = 1, x = y = 2, etc.) which will take some calculation but isn't hard and one can use the above stars and bars method to calculate each case. This number should be 81.
So (969 - 81)/2 = 444 is the number of non-negative integer solutions where x < y (since the number of solutions where x > y is exactly equal).
answered 2 hours ago
kyarykyary
583
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add a comment |
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$begingroup$
I can't comment so this answer is more of a comment. You use stars and bars. You have three bars | | | where each space represents one of x, y, w, or z. And you have 16 stars. So you count the number of distinct ways to rearrange the three bars and 16 stars, which is (3+16)!/(3! * 16!) = 969.
Next, find the number of cases where x = y (x = y = 1, x = y = 2, etc.) which will take some calculation but isn't hard and one can use the above stars and bars method to calculate each case. This number should be 81.
So (969 - 81)/2 = 444 is the number of non-negative integer solutions where x < y (since the number of solutions where x > y is exactly equal).
answered 2 hours ago
kyarykyary
583
$endgroup$
add a comment |
$begingroup$
I can't comment so this answer is more of a comment. You use stars and bars. You have three bars | | | where each space represents one of x, y, w, or z. And you have 16 stars. So you count the number of distinct ways to rearrange the three bars and 16 stars, which is (3+16)!/(3! * 16!) = 969.
Next, find the number of cases where x = y (x = y = 1, x = y = 2, etc.) which will take some calculation but isn't hard and one can use the above stars and bars method to calculate each case. This number should be 81.
So (969 - 81)/2 = 444 is the number of non-negative integer solutions where x < y (since the number of solutions where x > y is exactly equal).
answered 2 hours ago
kyarykyary
583
$endgroup$
add a comment |
$begingroup$
I can't comment so this answer is more of a comment. You use stars and bars. You have three bars | | | where each space represents one of x, y, w, or z. And you have 16 stars. So you count the number of distinct ways to rearrange the three bars and 16 stars, which is (3+16)!/(3! * 16!) = 969.
Next, find the number of cases where x = y (x = y = 1, x = y = 2, etc.) which will take some calculation but isn't hard and one can use the above stars and bars method to calculate each case. This number should be 81.
So (969 - 81)/2 = 444 is the number of non-negative integer solutions where x < y (since the number of solutions where x > y is exactly equal).
answered 2 hours ago
kyarykyary
583
$endgroup$
I can't comment so this answer is more of a comment. You use stars and bars. You have three bars | | | where each space represents one of x, y, w, or z. And you have 16 stars. So you count the number of distinct ways to rearrange the three bars and 16 stars, which is (3+16)!/(3! * 16!) = 969.
Next, find the number of cases where x = y (x = y = 1, x = y = 2, etc.) which will take some calculation but isn't hard and one can use the above stars and bars method to calculate each case. This number should be 81.
So (969 - 81)/2 = 444 is the number of non-negative integer solutions where x < y (since the number of solutions where x > y is exactly equal).
answered 2 hours ago
kyarykyary
583
answered 2 hours ago
kyarykyary
583
answered 2 hours ago
kyarykyary
583
answered 2 hours ago
kyarykyary
583
583
add a comment |
add a comment |
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3
$begingroup$
It's half the number of solutions for $xneq y$. If you find the number of solution for $x = y$, so the number of solutions of $2x + w + z =16$ then you just need to subtract that from the total number of solutions of $x+y+w+z$ and divide that by 2.
$endgroup$
– Count Iblis
2 hours ago
$begingroup$
Make that an answer!
$endgroup$
– Toby Mak
2 hours ago