How many solution for the equation $x+y=m+6$ $$ x+ y = m + 6 $$
$$ 1 \le x,y \le 6 $$
My calculations are definitely wrong.
I'm trying to solve the following equation instead:
$$ w_1+ w_2 = m + 4 $$
$$ w_1=x-1, w_2=y-2$$
$$ 0 \le w_1,w_2 \le 5 $$


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*I'm having a hard time to understand why the amount of solutions for the above equation is equal to the number of solutions for the first one.


Thirdly, to solve the above, I will calculated how many solutions for:
$$ w_1+ w_2 = m + 4 $$
$$ 6 \le w_1 (or) 6 \le w_2 $$
To calculate how many solutions for the above, I will calculate how many solutions for:
$$ w_1+ w_2 = m + 4 $$
$$ 0 \le w_1,w_2 $$
Which is $ \binom{m+5}{m+4} = m+5$. I solved it by ordering $(m+4)$ a and $1$ b in a line.
and how many solutions for:
$$ z_1+ w_2 = m - 2,w_1+ z_2 = m - 2,z_1+ z_2 = m - 8 $$
$$ z_1=w_1-6, z_2=w_2-6 $$
$$ 6 \le z_1,z_2,0 \le w_1,w_2 $$
The amount of solutions for $ z_1+ w_2 = m - 2 $ is $ m-1 $.
The amount of solutions for $ w_1+ z_2 = m - 2 $ is $ m-1 $.
The amount of solutions for $ z_1+ z_2  = m - 8 $ is $ m-7 $.
So the amount of solutions for $ w_1+ w_2 = m + 4 $ with the restrication of $ 6 \le w_1 or 6 \le w_2 $ is: (by the Inclusion–exclusion principle)
$$ m-1 + m-1 - (m-7) = m+5 $$
As you can notice, I'm definitely wrong because the amount of solutions to the same equation with and without restrictions is the same.
The final result for $ x+ y = m + 6 $ is $ 0 $ which is absolutly wrong.
 A: Hints
You mean to say $x\ge 1$,$y\le 6$ and $0\le m\le 5$
Then try breaking question into parts like $$x+y=6$$ in this case $m=0$ . Hence by star and bars method we get answer for this case as $6$.
Then try finding solutions for $m=1$ i.e. $$x+y=7$$ again by star and bars method we get solution as $7$ for this case. 
Similarly continue for $m=2,3,4,5$ and sum up the result to get final answer. 
A: Do u know that no. of "positive integer" solutions of the equation $x_1+x_2+...+x_n=m$ where $x_i$'s & $m$ is positive integer is $\binom{m-1}{n-1}$ 
So using this putting each $m$ from 0 to 5 , u get the number of solution and then sum it for total solution
A: So u have to subtract the case when $y\geq 7$ . 
Now let, $x+y=n$ where $x\geq 1$ and $y\geq 7$ 
now consider $x'=x-1\geq 0$ and $y'=x-7\geq 0$
so this gives $x'+y'=n-8$ so non-negative solution for this is $\binom{n-7}{1}$. 
This is it u have subtract each step. Let me Know when you finished.
A: Okay if you consider positive solutions , then answer should be $\displaystyle \sum_{m=0}^{5}\binom{m+5}{1}=45$
If you consider non-negative solutions , then number of solutions for some m is $\binom{n+m-1}{n-1}$, so there will be $\displaystyle \sum_{m=0}^{5}\binom{m+7}{1} = 57$
