My question is : How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 +x_5= 10$ where $x_1, x_2, x_3, x_4, x_5$ are positve integers and $x_1$ is an odd number?

I tried to solve it using Stars and bars, by getting to this formula $x_1=2y_1, x_2=y_2+1,x_3=y_3+1,x_4=y_4+1,x_5=y_5+1.$ which equals to $2y_1+y_2+y_3+y_4+y_5=6$. I don't know how to continue .

appreciate your help very much!

  • 5
    $\begingroup$ There are just $3$ cases, $x_1 \in \{1,3,5\}$. For each case, distribute the remaining stars to $x_2$ to $x_5$. $\endgroup$ – peterwhy Dec 29 '19 at 1:47

As mentioned in the comments, there are three cases: $x_1\in \{1,3,5\}$. That's because the minimum value of $x_2+x_3+x_4+x_5$, when all are positive integers, is $4$ and so the maximum odd value of $x_1$ is $5$. If we set $x_1 =1$, then there are $9$ stars left and $8$ spaces to place $3$ bars since all four remaining variables must be positive integers. This gives ${8\choose 3}=56$ possibilities.

If $x_1 = 3$, there are $6$ spaces for the bars and ${6\choose 3} = 20$ possibilities.

Finally, if $x_1 = 5$, there are $4$ spaces for the bars, for a total of ${4\choose 3} = 4$ possibilities.

Hence there are $56+20+4 = 80$ possibilities in total.


Required number of solutions is the coefficient of $x^{10}$ in the expansion of $$(x+x^3+x^5+x^7+\cdots)(x+x^2+x^3+x^4+\cdots)^4$$ $$=x^5(1+x^2+x^4+\cdots)(1+x+x^2+x^3+\cdots)^4$$ $$=\frac{x^5}{(1-x^2)(1-x)^4}$$ $$=\frac{x^5}{(1+x)(1-x)^5}$$ Now, expand both terms of denominator in power series expansion using formula and then find required coefficient.


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.