# 2016 AMC 10A Problems/Problem 20

## Problem

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 1

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 (also known as ball and urn) to see that there are $\binom{(N-4)+4}{4}$ or $\binom{N}{4}$ 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 $\boxed{\textbf{(B) }14.}$

Note: An alternative is instead of making the transformation, we "give" the variables $x, y, z, w$ 1, and then proceed as above.

~ Mathkiddie(minor edits by vadava_lx)

## Solution 2

By the Hockey Stick Identity, the number of terms that have all $a,b,c,d$ raised to a positive power is $\binom{N-1}{3}+\binom{N-2}{3}+\cdots + \binom{4}{3}+\binom{3}{3}=\binom{N}{4}$. We now want to find some $N$ such that $\binom{N}{4} = 1001$. As mentioned above, after noticing that $1001 = 7\cdot11\cdot13$, and some trial and error, we find that $\binom{14}{4} = 1001$, giving us our answer of $\boxed{\textbf{(B) }14.}$

~minor edits by vadava_lx

## Solution 3 (Casework)

The terms are in the form $a^xb^yc^zd^w1^t$, where $x + y + z + w + t = N$. The problem becomes distributing $N$ identical balls to $5$ different boxes $(x, y, z, w, t)$ such that each of the boxes $(x, y, z, w)$ has at least $1$ ball. The $N$ balls in a row have $N-1$ gaps among them. We are going to put $4$ or $3$ divisors into those $N-1$ gaps. There are $2$ cases of how to put the divisors.

Case $1$: Put 4 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d, 1)$ has at least one term. There are $\binom{N-1}{4}$ terms.

Case $2$: Put 3 divisors into $N-1$ gaps. It corresponds to each of $(a, b, c, d)$ has at least one term. There are $\binom{N-1}{3}$ terms.

So, there are $\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}$ terms. $\binom{N}{4} = 1001$, and since we have $\binom{14}{4} = 1001, N=\boxed{\textbf{(B) }14.}$

~ pi_is_3.14

## Video Solution 2

https://youtu.be/TpG8wlj4eRA with 5 Stars and Bars examples preceding the solution. Time stamps in description to skip straight to solution.

~IceMatrix

## See Also

 2016 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.