2016 AMC 10A Problems/Problem 20

For some particular value of $N$ , when $(a+b+c+d+1)^N$ is expanded and like terms are combined, the resulting expression contains exactly $1001$ terms that include all four variables $a, b,c,$ and $d$ , each to some positive power. What is $N$ ?

$\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution

All the desired terms are in the form $a^xb^yc^zd^w1^t$ , where $x + y + z + w + t = N$ (the $1^t$ part is necessary to make stars and bars work better.) Since $x$ , $y$ , $z$ , and $w$ must be at least $1$ ( $t$ can be $0$ ), let $x' = x - 1$ , $y' = y - 1$ , $z' = z - 1$ , and $w' = w - 1$ , so $x' + y' + z' + w' + t = N - 4$ . Now, we use stars and bars to see that there are $\binom{N}{4}$ solutions to this equation. We notice that $1001=7\cdot11\cdot13$ , which leads us to guess that $N$ is around these numbers. This suspicion proves to be correct, as we see that $\binom{14}{4} = 1001$ , giving us our answer of $\boxed{N=14}$ .

Another solution

The number of terms that have all $a,b,c,d$ raised to a positive power is $\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}$ . For $N=\boxed{14}$ , $\binom{N}{4}=1001$ .

2016 AMC 10A (Problems • Answer Key • Resources)
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2016 AMC 10A Problems/Problem 20

Solution

Another solution

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