How many cubic (i.e., third-degree) polynomials \(f(x)\) are there such that \(f(x)\) has nonnegative integer coefficients and \(f(1)=9\)?
Hint 1: Keep in mind that these are nonnegative integers not just postive integers.
Hint 2: Try to think about the problem by using "sticks and stones"
+526 Consider a third-degree polynomial \(f(x)\) such that
\(f(x) =ax^3+bx^2+cx+d\)
⇒ \(f(1) = a+b+c+d\)
Now, since \(f(1)=9\)
∴ \(a+b+c+d=9\)
So, you're looking for the number of ways to write 9 as the sum of four non-negative integers.
Here I'm applying "sticks and stones" method as you suggested.
⇒No. of solutions \(=\binom{n+r-1}{r-1}\)
Here, \(n=9\) and \(r=4\)
Total no. of possible solutions \(=\binom{9+4-1}{4}\)
\(=\binom{12}{4}\)
\(={12! \over (12-4)!4!}\)
\(={12! \over 8!4!}\)
\(=495\)
∴ There are 495 possible third-degree polynomials that fulfills the given condition.
Phew! This question blew my mind! 
P.S. Suggestions and corrections are welcome.
Nice solution, but I think that you have to consider that a cannot equal 0. (otherwise it wouldn't be cubic
)
+526 Yeah you're right, in that case lets eliminate one possibility and make it 494 possible polynomials.
