Combinatorics:No of solutions to an equation

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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.







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  • 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.
    – 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.
    – Sparsh
    Aug 24 at 5:33










  • I think I got it.Please let me know if my reasoning is correct.So
    – 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.
    – N. F. Taussig
    Aug 24 at 8:11














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down vote

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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.







share|cite|improve this question


















  • 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.
    – 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.
    – Sparsh
    Aug 24 at 5:33










  • I think I got it.Please let me know if my reasoning is correct.So
    – 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.
    – N. F. Taussig
    Aug 24 at 8:11












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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.







share|cite|improve this question














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.









share|cite|improve this question













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edited Aug 24 at 6:56









postmortes

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1,56311016










asked Aug 24 at 4:42









Sparsh

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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.
    – 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.
    – Sparsh
    Aug 24 at 5:33










  • I think I got it.Please let me know if my reasoning is correct.So
    – 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.
    – N. F. Taussig
    Aug 24 at 8:11












  • 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.
    – 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.
    – Sparsh
    Aug 24 at 5:33










  • I think I got it.Please let me know if my reasoning is correct.So
    – 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.
    – 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.
– 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.
– 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.
– 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.
– Sparsh
Aug 24 at 5:33












I think I got it.Please let me know if my reasoning is correct.So
– Sparsh
Aug 24 at 5:42




I think I got it.Please let me know if my reasoning is correct.So
– 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.
– 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.
– N. F. Taussig
Aug 24 at 8:11















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