How many cubic (i.e., third-degree) polynomials f(x) are there such that f(x) has nonnegative integer coefficients and f(1) = 9?
f(1) = 9 means ax^3 + bx^2 + cx + d = 9. But we can have coefficients of 0. I am really confused.
By stars and bars, the number of such polynomials is C(12,4) = 495.
Guest, you are super close, but note that a cannot be 0 as that will violate the definition of a cubic polynomial.
By f(1) = 9, you will get that:
a + b + c + d = 9
Since a has to be greater than or equal to 1, we can set another variable f that is equivalent to a - 1.
f + b + c + d = 8
We can use stars and bars from here, as f can take a value of 0 as well. Hence, the answer is actually
C(11, 3) = 11 x 10 x 9/6 = 15 x 11 = 165.