Combinatorics:No of solutions to an equation
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Find the number of solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 22$, where $x_1, x_2, x_3, x_4$ and $x_5$ are non-negative integers, and $x_1+x_2 leq 2$. (You may leave the answer as an expression consisting of binomial coefficients.)
So I tried solving it and I think we need to consider the cases when $x_1+x_2$ is equal to $2$; when $x_1+x_2$ is equal to $1$ and so on.
combinatorics
add a comment |Â
up vote
1
down vote
favorite
Find the number of solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 22$, where $x_1, x_2, x_3, x_4$ and $x_5$ are non-negative integers, and $x_1+x_2 leq 2$. (You may leave the answer as an expression consisting of binomial coefficients.)
So I tried solving it and I think we need to consider the cases when $x_1+x_2$ is equal to $2$; when $x_1+x_2$ is equal to $1$ and so on.
combinatorics
1
Your thought is correct. You have three problems, one for each sum of $x1+x2$. Are you familiar with stars and bars? Search the site or Wikipedia for it.
â Ross Millikan
Aug 24 at 5:22
Yes but the thing I want to know is is this using principle of inclusion and exclusion here.The answer is 24C2 +2*23C2 +3*22C2.I am not sure why they are multiplying by 2 and 3,Could you elaborate please.
â Sparsh
Aug 24 at 5:33
I think I got it.Please let me know if my reasoning is correct.So
â Sparsh
Aug 24 at 5:42
The answer you stated is correct. It is obtained by treating the cases $x_1 + x_2 = 0$, $x_1 + x_2 = 1$, and $x_1 + x_2 = 2$ separately, then adding the results.
â N. F. Taussig
Aug 24 at 8:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the number of solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 22$, where $x_1, x_2, x_3, x_4$ and $x_5$ are non-negative integers, and $x_1+x_2 leq 2$. (You may leave the answer as an expression consisting of binomial coefficients.)
So I tried solving it and I think we need to consider the cases when $x_1+x_2$ is equal to $2$; when $x_1+x_2$ is equal to $1$ and so on.
combinatorics
Find the number of solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 22$, where $x_1, x_2, x_3, x_4$ and $x_5$ are non-negative integers, and $x_1+x_2 leq 2$. (You may leave the answer as an expression consisting of binomial coefficients.)
So I tried solving it and I think we need to consider the cases when $x_1+x_2$ is equal to $2$; when $x_1+x_2$ is equal to $1$ and so on.
combinatorics
edited Aug 24 at 6:56
postmortes
1,56311016
1,56311016
asked Aug 24 at 4:42
Sparsh
142
142
1
Your thought is correct. You have three problems, one for each sum of $x1+x2$. Are you familiar with stars and bars? Search the site or Wikipedia for it.
â Ross Millikan
Aug 24 at 5:22
Yes but the thing I want to know is is this using principle of inclusion and exclusion here.The answer is 24C2 +2*23C2 +3*22C2.I am not sure why they are multiplying by 2 and 3,Could you elaborate please.
â Sparsh
Aug 24 at 5:33
I think I got it.Please let me know if my reasoning is correct.So
â Sparsh
Aug 24 at 5:42
The answer you stated is correct. It is obtained by treating the cases $x_1 + x_2 = 0$, $x_1 + x_2 = 1$, and $x_1 + x_2 = 2$ separately, then adding the results.
â N. F. Taussig
Aug 24 at 8:11
add a comment |Â
1
Your thought is correct. You have three problems, one for each sum of $x1+x2$. Are you familiar with stars and bars? Search the site or Wikipedia for it.
â Ross Millikan
Aug 24 at 5:22
Yes but the thing I want to know is is this using principle of inclusion and exclusion here.The answer is 24C2 +2*23C2 +3*22C2.I am not sure why they are multiplying by 2 and 3,Could you elaborate please.
â Sparsh
Aug 24 at 5:33
I think I got it.Please let me know if my reasoning is correct.So
â Sparsh
Aug 24 at 5:42
The answer you stated is correct. It is obtained by treating the cases $x_1 + x_2 = 0$, $x_1 + x_2 = 1$, and $x_1 + x_2 = 2$ separately, then adding the results.
â N. F. Taussig
Aug 24 at 8:11
1
1
Your thought is correct. You have three problems, one for each sum of $x1+x2$. Are you familiar with stars and bars? Search the site or Wikipedia for it.
â Ross Millikan
Aug 24 at 5:22
Your thought is correct. You have three problems, one for each sum of $x1+x2$. Are you familiar with stars and bars? Search the site or Wikipedia for it.
â Ross Millikan
Aug 24 at 5:22
Yes but the thing I want to know is is this using principle of inclusion and exclusion here.The answer is 24C2 +2*23C2 +3*22C2.I am not sure why they are multiplying by 2 and 3,Could you elaborate please.
â Sparsh
Aug 24 at 5:33
Yes but the thing I want to know is is this using principle of inclusion and exclusion here.The answer is 24C2 +2*23C2 +3*22C2.I am not sure why they are multiplying by 2 and 3,Could you elaborate please.
â Sparsh
Aug 24 at 5:33
I think I got it.Please let me know if my reasoning is correct.So
â Sparsh
Aug 24 at 5:42
I think I got it.Please let me know if my reasoning is correct.So
â Sparsh
Aug 24 at 5:42
The answer you stated is correct. It is obtained by treating the cases $x_1 + x_2 = 0$, $x_1 + x_2 = 1$, and $x_1 + x_2 = 2$ separately, then adding the results.
â N. F. Taussig
Aug 24 at 8:11
The answer you stated is correct. It is obtained by treating the cases $x_1 + x_2 = 0$, $x_1 + x_2 = 1$, and $x_1 + x_2 = 2$ separately, then adding the results.
â N. F. Taussig
Aug 24 at 8:11
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2892784%2fcombinatoricsno-of-solutions-to-an-equation%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Your thought is correct. You have three problems, one for each sum of $x1+x2$. Are you familiar with stars and bars? Search the site or Wikipedia for it.
â Ross Millikan
Aug 24 at 5:22
Yes but the thing I want to know is is this using principle of inclusion and exclusion here.The answer is 24C2 +2*23C2 +3*22C2.I am not sure why they are multiplying by 2 and 3,Could you elaborate please.
â Sparsh
Aug 24 at 5:33
I think I got it.Please let me know if my reasoning is correct.So
â Sparsh
Aug 24 at 5:42
The answer you stated is correct. It is obtained by treating the cases $x_1 + x_2 = 0$, $x_1 + x_2 = 1$, and $x_1 + x_2 = 2$ separately, then adding the results.
â N. F. Taussig
Aug 24 at 8:11