2016 AMC 10A Problem 20

After combining like terms, each distinct term is $a^i b^j c^k d^l$ where the exponents record how many of the $N$ factors contributed $a$, $b$, $c$, $d$, or $1$. If $m$ factors contributed the $1$, then
$$i + j + k + l + m = N, \qquad m \ge 0.$$

We want the terms with $i, j, k, l \ge 1$. Give each of $a, b, c, d$ one forced unit: set $i' = i-1$, etc. Then $i' + j' + k' + l' + m = N - 4$ with all five variables nonnegative. By stars and bars the number of solutions is
$$\binom{(N-4) + 4}{4} = \binom{N}{4}.$$

Set $\binom{N}{4} = 1001 = 7 \cdot 11 \cdot 13$. Since $\binom{14}{4} = \frac{14\cdot 13\cdot 12\cdot 11}{24} = 1001$, we get $N = 14$. (The function $\binom{N}{4}$ is increasing, so this is the only solution.)

The answer is $\boxed{\textbf{(B)}\ 14}$.

Counting distinct monomials in an expansion is counting exponent tuples, and the constant term $1$ acts as a fifth variable that absorbs whatever degree is left over. Forcing positive exponents is the standard "hand out one unit first" reduction to nonnegative stars and bars. The factorization $1001 = 7\cdot 11\cdot 13$ is a strong hint toward $\binom{14}{4}$.

Test the choices directly with $\binom{N}{4}$: $\binom{9}{4} = 126$, $\binom{14}{4} = 1001$, $\binom{16}{4} = 1820$. Only $N = 14$ works.

• Dropping the constant 1 from the count (treating the exponents as summing to exactly N with no slack), which gives C(N-1, 3) and no matching choice.
• Also forcing the exponent of $1$ to be positive, which yields $\binom{N-1}{4}$ and $N = 15$, not among the choices.