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How many are the non-negative integer solutions of $𝑥 + 𝑦 + 𝑤 + 𝑧 = 16$ where $x

Homework - How many non negative solutions?Number of non-negative integers solutions of $x_1 + x_2 + x_3 + x_4 + x_5 = 10$ when $x_1 = x_2$ and when $x_1 > x_2$How many distinct, non-negative integer solutions are there for $2x_0+sum_i=1^mx_i=n$?How many non negative integer solutions are there for the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 36$ with restrictions?How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 leq 35$ in which all the $x_i$ are non-negative integers?How many integer-valued solutions are there?How many non-negative integer solutions are there to $sum_i=1^m ia_i=n$?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 $-10leq x_ileq20$?

1

$begingroup$

How many are the non-negative integer solutions of $𝑥 + 𝑦 + 𝑤 + 𝑧 = 16$ where $x < y$?

Anyone can explain how to think to approach this type of problem?
The answer is 444.

combinatorics discrete-mathematics permutations

share|cite|improve this question

edited May 12 at 4:21

YuiTo Cheng

3,26371345

asked May 12 at 1:17

gmn_1450gmn_1450

406

$endgroup$

closed as off-topic by YuiTo Cheng, Jean-Claude Arbaut, Cesareo, Lee David Chung Lin, Jendrik Stelzner May 12 at 19:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YuiTo Cheng, Jean-Claude Arbaut, Cesareo, Lee David Chung Lin, Jendrik Stelzner

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  May 12 at 1:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make that an answer!
  $endgroup$
  – Toby Mak
  May 12 at 1:24
  

  
  
  
  

add a comment | 

1

$begingroup$

How many are the non-negative integer solutions of $𝑥 + 𝑦 + 𝑤 + 𝑧 = 16$ where $x < y$?

Anyone can explain how to think to approach this type of problem?
The answer is 444.

combinatorics discrete-mathematics permutations

share|cite|improve this question

edited May 12 at 4:21

YuiTo Cheng

3,26371345

asked May 12 at 1:17

gmn_1450gmn_1450

406

$endgroup$

closed as off-topic by YuiTo Cheng, Jean-Claude Arbaut, Cesareo, Lee David Chung Lin, Jendrik Stelzner May 12 at 19:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – YuiTo Cheng, Jean-Claude Arbaut, Cesareo, Lee David Chung Lin, Jendrik Stelzner

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  May 12 at 1:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make that an answer!
  $endgroup$
  – Toby Mak
  May 12 at 1:24
  

  
  
  
  

add a comment | 

$begingroup$

How many are the non-negative integer solutions of $𝑥 + 𝑦 + 𝑤 + 𝑧 = 16$ where $x < y$?

Anyone can explain how to think to approach this type of problem?
The answer is 444.

combinatorics discrete-mathematics permutations

share|cite|improve this question

edited May 12 at 4:21

YuiTo Cheng

3,26371345

asked May 12 at 1:17

gmn_1450gmn_1450

406

$endgroup$

share|cite|improve this question

edited May 12 at 4:21

YuiTo Cheng

3,26371345

asked May 12 at 1:17

gmn_1450gmn_1450

406

share|cite|improve this question

edited May 12 at 4:21

YuiTo Cheng

3,26371345

asked May 12 at 1:17

gmn_1450gmn_1450

406

edited May 12 at 4:21

YuiTo Cheng

3,26371345

edited May 12 at 4:21

YuiTo Cheng

3,26371345

asked May 12 at 1:17

gmn_1450gmn_1450

406

asked May 12 at 1:17

gmn_1450gmn_1450

406

2 Answers
2

active

oldest

votes

7

$begingroup$

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).

share|cite|improve this answer

answered May 12 at 1:26

kyarykyary

783

$endgroup$

add a comment | 

0

$begingroup$

Consider cases and find a pattern.

When $x=0$:
$$0+y+z+w=16, yge 1, z,wge 0 Rightarrow \
t+1+z+w=16,t,z,wge 0 Rightarrow \
t+z+w=15, t,z,wge 0 Rightarrow \
15+3-1choose 3-1$$
Note: It was used Stars and Bars.

When $x=1$:
$$1+y+z+w=16, yge 2, z,wge 0 Rightarrow \
t+2+z+w=15,t,z,wge 0 Rightarrow \
t+z+w=13, t,z,wge 0 Rightarrow \
13+3-1choose 3-1$$
$vdots$

When $x=7$:
$$7+y+z+w=16, yge 8, z,wge 0 Rightarrow \
t+8+z+w=9,t,z,wge 0 Rightarrow \
t+z+w=1, t,z,wge 0 Rightarrow \
1+3-1choose 3-1$$
Hence, it is the sum:
$$beginalignsum_i=1^8 2i-1+3-1choose 3-1&=sum_i=1^8 2i+1choose 2=sum_i=1^8 frac(2i+1)!2!(2i-1)!=sum_i=1^8 frac(2i+1)2i2=\
&=sum_i=1^8 (2i^2+i)=2frac8(8+1)(2cdot 8+1)6+frac8(8+1)2=\
&=24cdot 17+36=444.endalign$$

share|cite|improve this answer

answered May 12 at 6:30

farruhotafarruhota

23.2k2942

$endgroup$

add a comment | 

7

$begingroup$

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).

share|cite|improve this answer

answered May 12 at 1:26

kyarykyary

783

$endgroup$

add a comment | 

7

$begingroup$

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).

share|cite|improve this answer

answered May 12 at 1:26

kyarykyary

783

$endgroup$

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).

share|cite|improve this answer

answered May 12 at 1:26

kyarykyary

783

$endgroup$

share|cite|improve this answer

answered May 12 at 1:26

kyarykyary

783

answered May 12 at 1:26

kyarykyary

783

answered May 12 at 1:26

kyarykyary

783

0

$begingroup$

Consider cases and find a pattern.

When $x=0$:
$$0+y+z+w=16, yge 1, z,wge 0 Rightarrow \
t+1+z+w=16,t,z,wge 0 Rightarrow \
t+z+w=15, t,z,wge 0 Rightarrow \
15+3-1choose 3-1$$
Note: It was used Stars and Bars.

When $x=1$:
$$1+y+z+w=16, yge 2, z,wge 0 Rightarrow \
t+2+z+w=15,t,z,wge 0 Rightarrow \
t+z+w=13, t,z,wge 0 Rightarrow \
13+3-1choose 3-1$$
$vdots$

When $x=7$:
$$7+y+z+w=16, yge 8, z,wge 0 Rightarrow \
t+8+z+w=9,t,z,wge 0 Rightarrow \
t+z+w=1, t,z,wge 0 Rightarrow \
1+3-1choose 3-1$$
Hence, it is the sum:
$$beginalignsum_i=1^8 2i-1+3-1choose 3-1&=sum_i=1^8 2i+1choose 2=sum_i=1^8 frac(2i+1)!2!(2i-1)!=sum_i=1^8 frac(2i+1)2i2=\
&=sum_i=1^8 (2i^2+i)=2frac8(8+1)(2cdot 8+1)6+frac8(8+1)2=\
&=24cdot 17+36=444.endalign$$

share|cite|improve this answer

answered May 12 at 6:30

farruhotafarruhota

23.2k2942

$endgroup$

add a comment | 

0

$begingroup$

Consider cases and find a pattern.

When $x=0$:
$$0+y+z+w=16, yge 1, z,wge 0 Rightarrow \
t+1+z+w=16,t,z,wge 0 Rightarrow \
t+z+w=15, t,z,wge 0 Rightarrow \
15+3-1choose 3-1$$
Note: It was used Stars and Bars.

When $x=1$:
$$1+y+z+w=16, yge 2, z,wge 0 Rightarrow \
t+2+z+w=15,t,z,wge 0 Rightarrow \
t+z+w=13, t,z,wge 0 Rightarrow \
13+3-1choose 3-1$$
$vdots$

When $x=7$:
$$7+y+z+w=16, yge 8, z,wge 0 Rightarrow \
t+8+z+w=9,t,z,wge 0 Rightarrow \
t+z+w=1, t,z,wge 0 Rightarrow \
1+3-1choose 3-1$$
Hence, it is the sum:
$$beginalignsum_i=1^8 2i-1+3-1choose 3-1&=sum_i=1^8 2i+1choose 2=sum_i=1^8 frac(2i+1)!2!(2i-1)!=sum_i=1^8 frac(2i+1)2i2=\
&=sum_i=1^8 (2i^2+i)=2frac8(8+1)(2cdot 8+1)6+frac8(8+1)2=\
&=24cdot 17+36=444.endalign$$

share|cite|improve this answer

answered May 12 at 6:30

farruhotafarruhota

23.2k2942

$endgroup$

add a comment | 

$begingroup$

Consider cases and find a pattern.

When $x=0$:
$$0+y+z+w=16, yge 1, z,wge 0 Rightarrow \
t+1+z+w=16,t,z,wge 0 Rightarrow \
t+z+w=15, t,z,wge 0 Rightarrow \
15+3-1choose 3-1$$
Note: It was used Stars and Bars.

When $x=1$:
$$1+y+z+w=16, yge 2, z,wge 0 Rightarrow \
t+2+z+w=15,t,z,wge 0 Rightarrow \
t+z+w=13, t,z,wge 0 Rightarrow \
13+3-1choose 3-1$$
$vdots$

When $x=7$:
$$7+y+z+w=16, yge 8, z,wge 0 Rightarrow \
t+8+z+w=9,t,z,wge 0 Rightarrow \
t+z+w=1, t,z,wge 0 Rightarrow \
1+3-1choose 3-1$$
Hence, it is the sum:
$$beginalignsum_i=1^8 2i-1+3-1choose 3-1&=sum_i=1^8 2i+1choose 2=sum_i=1^8 frac(2i+1)!2!(2i-1)!=sum_i=1^8 frac(2i+1)2i2=\
&=sum_i=1^8 (2i^2+i)=2frac8(8+1)(2cdot 8+1)6+frac8(8+1)2=\
&=24cdot 17+36=444.endalign$$

share|cite|improve this answer

answered May 12 at 6:30

farruhotafarruhota

23.2k2942

$endgroup$

share|cite|improve this answer

answered May 12 at 6:30

farruhotafarruhota

23.2k2942

answered May 12 at 6:30

farruhotafarruhota

23.2k2942

answered May 12 at 6:30

farruhotafarruhota

23.2k2942