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Problem:How many cubic (i.e., third-degree) polynomials $f(x)$ are there such that $f(x)$ has nonnegative integer coefficients and $f(1)=9$?

My take: Let the polynomial be $ax^3+bx^2+cx+d$ which means $f(1)=a+b+c+d = 9.$ Therefore the number of solutions is $\binom{12}{3}=220.$

But however, this is wrong, so where did I go wrong?

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    $\begingroup$ You must make sure that a is always greater than or equal to 1 for the polynomial to be cubic $\endgroup$ – Samyak Jha Jul 28 '20 at 7:31
  • $\begingroup$ So I let $a' = a+1, b' = b+1, c' = c+1$ which means that a' will never be negative. $\endgroup$ – Michael Li Jul 28 '20 at 7:56
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    $\begingroup$ No, just $a'=a-1$........... $\endgroup$ – Aqua Jul 28 '20 at 8:26
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If $a=0$ is allowed, your formula is correct. Since you say that it's wrong, it seems that it must be $a>0$. Therefore, you are counting the number of ways to choose $4$ positive integers $a$, $b+1$, $c+1$ and $d+1$ such that their sum is $a+(b+1)+(c+1)+(d+1)=9+3=12$. This is equivalent to placing $3$ bars at the $11$ different places between $12$ stars, so the number you are looking for is $$\binom{11}{3}=165$$

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