Number of solutions to $x_1 + \ldots + x_5 = 26$ for $x_k \leq 10$ Find the number of solutions to $x_1 + \ldots + x_5 = 26$ for $x_k \leq 10$  
The answer is
$$\binom{30}4 - \binom{5}1 \binom{19}4 + \binom{5}2\binom{8}4$$  
by Inclusion-Exclusion principle.  
However, why can't I just simply use the substitution $y_k = 10-x_j \geq 0$
to count number of solutions to
$$y_1 + \ldots + y_5 = 50 - 26 = 24$$ 
for $y_k \geq 0$?  
This would be
$$\binom{28}4.$$
 A: When restrictions like $x_i\leq\square$ or $x_i\geq\square$  are encountered we should subtract the opposite situations from all the results, but we have to be careful about the maximal amount of numbers we can equate to $10$ ,here it is $2$  we will evaluate it firstly for each $1$ $x_i$ and then two different $x_i$ $x_j$ ; the whole situations are:$$\dbinom{26+5-1}{5-1}$$ and the situations in which $x_i\geq 11$ are to be subtracted: $$x_1'+11+x_2+\cdots+x_5=26$$ from here $$5\dbinom{15+5-1}{5-1}$$ (we added the $5$ because we substitude $11$ for each $x_i$ and now we do it for $2$;$$x_1'+11+x_2'+11+x_3+\cdots+x_5=26$$ that is; $$\dbinom{5}{2}\dbinom{4+5-1}{5-1}$$ and now we bring them together with; $$\dbinom{30}{4}-5\dbinom{19}{4}+\dbinom{5}{2}\dbinom{8}{4}$$ (This $+$ and $-$ thing is a principle driver from sets $s(A\cup B)=s(A)+s(B)-s(A\cap B)$ we have multiplied a $-$ up there)(P.S I have used that the integer solutions for $x_1+\cdots+x_r=n$ , $\dbinom{n+r-1}{r-1}$
A: Since 
$$
0\leq x_j\leq 10
$$
then
$$
10\geq10-x_j\geq0
$$
which implies that the substitution does not change anything.
