2016 AMC 10A Problems/Problem 20

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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 $\binom{14}{4} = 1001$, partly because 1001=7 * 11 * 13 which gives us some suspicions, so $N = \boxed{14}$.

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 $N=\boxed{14}$, $\binom{N}{4}=1001$.

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions

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