Find the number of integer solutions to $x_1+x_2+x_3+x_4= 30$ where $0\leq x_n <10$ for $1\leq n \leq 4$ I'm reviewing for my final coming up and I'm not sure about my solution to this problem.

Find the number of integer solutions to $$x_1+x_2+x_3+x_4= 30$$ where $0\leq x_n <10$ for $1\leq n \leq 4$.

My first thought was to find the number of solutions where at least one of the x's is greater than ten and subtracting that from the total, which is equivalent to finding the number of integer solutions to 
Find the number of integer solutions to $$x_1+x_2+x_3+x_4= 20.$$
That's a problem I know how to solve, and the answer is $nCr(23,20)=1771$.
Since there are $nCr(33,30)=5456$ total solutions, the final answer is $$5456-1771=3685.$$
Is this right?
 A: $$|s|=\binom{30+4-1}{4-1}=\binom{33}{3}=\binom{33}{3}\\
A :x_1 \geq10 \to (x_1-10)+x_2+x_3+x_4=30-10 \to |A|=\binom{20+4-1}{4-1}=\binom{23}{3}\\
B::x_2 \geq10 \to x_1+(x_2-10)+x_3+x_4=30-10 \to |B|=\binom{23}{3}\\
C::x_3 \geq10 \to x_1+x_2+(x_3-10)+x_4=30-10 \to |C|=\binom{23}{3}\\
C::x_4 \geq10 \to x_1+x_2+x_3+(x_4-10)=30-10 \to |D|=\binom{23}{3}\\
A\cap B:x_1 \geq 10 ,x_2 \geq 10 \to (x_1-10)+(x_2-10)+x_3+x_4=30-10-10 \to |A\cap B|=\binom{10+4-1}{4-1}=\binom{13}{3}\\
...
|A\cap C|=|A\cap D|=|B\cap C|=|B\cap D|=|C\cap D|=\binom{13}{3}\\
A \cap B\cap C:x_1,x_2,x_3 \geq10 \to (x_1-10)+(x_2-10)+(x_3-10)+x_4=30-30 \\ \to |A \cap B\cap C|=\binom{0+4-1}{4-1}=1\\|A \cap B\cap D|=|B \cap C\cap D|=|A \cap C\cap D|=\binom{3}{3}=1\\
A \cap B\cap C\cap D:x_1,x_2,x_3,x-4 \geq10 \to (x_1-10)+(x_2-10)+(x_3-10)+(x_4-10)=30-40 \to |A \cap B\cap C\cap D|=0 \\ $$now answer is 
$$\\{\color{Red}{|s|-|A\cup B\cup C\cup D|=\\|s|-((|A|+|B|+|C|+|D|)-(|A \cap B|+...)+(|A \cap B \cap C|+...)-(|A \cap B\cap C\cap D|))=\\
\binom{33}{3}-(4\binom{23}{3}-6\binom{13}{3}+4\binom{3}{3}-0)} } $$
A: HINT:
I got $$
(x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)^4 = \\
x^{36}+4 x^{35}+10 x^{34}+20 x^{33}+35 x^{32}+56 x^{31}+84 x^{30}+120 x^{29}+165 x^{28}+220 x^{27}+282 x^{26}+348 x^{25}+415 x^{24}+480 x^{23}+540 x^{22}+592 x^{21}+633 x^{20}+660 x^{19}+670 x^{18}+660 x^{17}+633 x^{16}+592 x^{15}+540 x^{14}+480 x^{13}+415 x^{12}+348 x^{11}+282 x^{10}+220 x^9+165 x^8+120 x^7+84 x^6+56 x^5+35 x^4+20 x^3+10 x^2+4 x+1$$
( cf. hint of @dREaM: ) so the number is $84$. 
A: It is correct. Answer is indeed  correct $nCr(33,3)=nCr(30,30)=5456$ . Here application of multinomial theorem is done , for solution to $x_1 + x_2 +x_3 +.....+ x_r = n$, there are $(n+r-1)C(r-1)$ [ which is also the number of terms in multinomial sum, is equal to the number of monomials of degree n on the variables $x_1, …, x_m$  and here n=30 , r=4 so we get the total possible solution as $nCr(33,3)=nCr(30,30)=5456$
