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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?

user26857
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Zachary F
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    No, that is no the number of solutions in which at least one $x$ is greater than $10$. – Asinomás Dec 10 '15 at 17:25
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    You could use inclusion-exclusion, or alternatively, what you want is the coefficient of $x^{30}$ in $(\frac{x^{11}-1}{x-1})^4$ – Asinomás Dec 10 '15 at 17:27
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    @dREaM: Minor typo, it should be $\left(\frac{x^{10}-1}{x-1}\right )^ 4$, great idea nevertheless! – orangeskid Dec 10 '15 at 18:32
  • Oh yeah, for some reason I thought $x_n\leq 10$. Good eye! – Asinomás Dec 10 '15 at 19:26
  • Why that function and that coefficient? – Zachary F Dec 10 '15 at 19:28
  • well, first of all $1+x+x^2+\dots +x^{9}=\frac{x^{10}-1}{x-1}$. Now, think of how you multiply expressions like these. you just have to take the sum over all possible combinations, (there are $10^4$ such combinations). So the coefficient of $x^{30}$ is equal to the number of combinations $x^{a_1}x^{a_2}x^{a_3}x^{a_4}$ with $a_1+a_2+a_2+a_4=30$. Where $a_n, 1\leq n \leq 4$ is one of the summands of the nth factor. So $0\leq a_n<10$. Which is exactly what you want to count. – Asinomás Dec 10 '15 at 19:34
  • Oh! So it's all the ways you can add up to 30 with 1 through 9. That's clever. And if I wanted it to be with, say, just even numbers I could do $(1+x^2+x^4+x^6+...+x^{30})^4$? – Zachary F Dec 10 '15 at 21:22

3 Answers3

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$$|s|=\binom{30+4-1}{4-1}=\binom{33}{3}=\binom{33}{3}\\ A :x_1 \geq10 \to (x_1-10)+x_2+x_3+x_4=30-10 \to |A|=\binom{20+4-1}{4-1}=\binom{23}{3}\\ B::x_2 \geq10 \to x_1+(x_2-10)+x_3+x_4=30-10 \to |B|=\binom{23}{3}\\ C::x_3 \geq10 \to x_1+x_2+(x_3-10)+x_4=30-10 \to |C|=\binom{23}{3}\\ C::x_4 \geq10 \to x_1+x_2+x_3+(x_4-10)=30-10 \to |D|=\binom{23}{3}\\ A\cap B:x_1 \geq 10 ,x_2 \geq 10 \to (x_1-10)+(x_2-10)+x_3+x_4=30-10-10 \to |A\cap B|=\binom{10+4-1}{4-1}=\binom{13}{3}\\ ... |A\cap C|=|A\cap D|=|B\cap C|=|B\cap D|=|C\cap D|=\binom{13}{3}\\ A \cap B\cap C:x_1,x_2,x_3 \geq10 \to (x_1-10)+(x_2-10)+(x_3-10)+x_4=30-30 \\ \to |A \cap B\cap C|=\binom{0+4-1}{4-1}=1\\|A \cap B\cap D|=|B \cap C\cap D|=|A \cap C\cap D|=\binom{3}{3}=1\\ A \cap B\cap C\cap D:x_1,x_2,x_3,x-4 \geq10 \to (x_1-10)+(x_2-10)+(x_3-10)+(x_4-10)=30-40 \to |A \cap B\cap C\cap D|=0 \\ $$now answer is $$\\{\color{Red}{|s|-|A\cup B\cup C\cup D|=\\|s|-((|A|+|B|+|C|+|D|)-(|A \cap B|+...)+(|A \cap B \cap C|+...)-(|A \cap B\cap C\cap D|))=\\ \binom{33}{3}-(4\binom{23}{3}-6\binom{13}{3}+4\binom{3}{3}-0)} } $$

Khosrotash
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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 $84$.

orangeskid
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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 $nCr(33,3)=nCr(30,30)=5456$

Murtuza Vadharia
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