How many solutions are there in the following equation over the natural numbers such that $x_1+x_2+x_3+x_4+x_5+x_6+x_7=30$ if $x_1+x_2+x_3>x_4+x_5+x_6+x_7$ ?

I made up a combinatorics question in my mind , but i stuck in it. What i tried is :

Firstly , i wanted to use symmetry property but , it has odd number of variables . Hence ,i could not do anything.

Secondly , i thought about whether generating functions can be used or not , but i stuck in writing the desired generating formula for it.

So , i hope to find nice approaches to my question.

$NOTE=$Natural numbers start with zero.

  • 2
    $\begingroup$ @BrianMScott last line implies that $0\in\mathbb N$. $\endgroup$ Mar 13, 2021 at 21:09

1 Answer 1


Using stars and bars you can write the number of solution as $$ \sum_{i=0}^{14}\binom{i+3}3\binom {30-i+2}2, $$ where the first factor counts the number of solutions to the equation $x_4+x_5+x_6+x_7=i $ and the second one those of $x_1+x_2+x_3=30-i $.

  • $\begingroup$ thanks for your answer , it is correct but , i hope to find easier way than you did.Because , for large numbers , this solution will be cumbersome process. By the way , i glad to see you again:) +1 $\endgroup$ Mar 13, 2021 at 21:26
  • $\begingroup$ @Bulbasaur Have you found an easier way? $\endgroup$
    – user
    Apr 17, 2021 at 21:42
  • $\begingroup$ I think that an elegant generating function can be found to get rid of cumbersome process and I appreciate your work for me. By the way , i forgot this question. I guess the new answer wont be come. If I can find nice way , i will say you. Thanks for your work.At last , lets forget the past bad experience between us. I appreciate your knowledge and good wills. Have a nice day.. $\endgroup$ Apr 18, 2021 at 10:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.