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? 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! 
 A: 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.
A: 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.
