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Difference between revisions of "2016 AMC 10A Problems/Problem 20"

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==Solution 3 (Stars and Bars)==
 
==Solution 3 (Stars and Bars)==
 
5 sections (<math>x,y,z,w,1</math>) 4 of which need at least 1 object. <math>\dbinom{N+4-4\cdot1}{4}</math>. Test the choices and find that <math>N=\boxed{\text{(B) }14}</math>
 
5 sections (<math>x,y,z,w,1</math>) 4 of which need at least 1 object. <math>\dbinom{N+4-4\cdot1}{4}</math>. Test the choices and find that <math>N=\boxed{\text{(B) }14}</math>
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==Solution 4 (Casework)==
 +
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The terms are in the form <math>a^xb^yc^zd^w1^t</math>, where <math>x + y + z + w + t = N</math>. The problem becomes distributing N identical balls to 5 different boxes <math>(x, y, z, w, t)</math>. There are <math>2</math> cases of how we count terms with <math>(a, b, c, d)</math>.
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Case <math>1</math>:
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Count the terms that have at least <math>1</math> of <math>(a, b, c, d, 1)</math>. There are <math>\binom{N-1}{4}</math> terms.
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Case <math>2</math>:
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Count the terms that have at least <math>1</math> of <math>(a, b, c, d)</math> but not <math>1</math>. There are <math>\binom{N-1}{3}</math> terms.
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So, there are <math>\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}=1001</math> terms. Similar to solutions <math>1, 2, 3</math>, we find <math>N=\boxed{\text{(B) }14}</math>
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~isabelchen
  
 
==Video Solution==
 
==Video Solution==

Revision as of 08:40, 5 October 2021

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 $N=\boxed{14}$.

  • An alternative is to instead make the transformation $t'=t+1$, so $x + y + z + w + t' = N + 1$, and all variables are positive integers. The solution to this, by Stars and Bars is $\binom{(N+1)-1}{4}=\binom{N}{4}$ and we can proceed as above.

Solution 2

By 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 $N=\boxed{14}$

Solution 3 (Stars and Bars)

5 sections ($x,y,z,w,1$) 4 of which need at least 1 object. $\dbinom{N+4-4\cdot1}{4}$. Test the choices and find that $N=\boxed{\text{(B) }14}$

Solution 4 (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 $(x, y, z, w, t)$. There are $2$ cases of how we count terms with $(a, b, c, d)$.

Case $1$: Count the terms that have at least $1$ of $(a, b, c, d, 1)$. There are $\binom{N-1}{4}$ terms.

Case $2$: Count the terms that have at least $1$ of $(a, b, c, d)$ but not $1$. There are $\binom{N-1}{3}$ terms.

So, there are $\binom{N-1}{4}+\binom{N-1}{3}=\binom{N}{4}=1001$ terms. Similar to solutions $1, 2, 3$, we find $N=\boxed{\text{(B) }14}$

~isabelchen

Video Solution

https://www.youtube.com/watch?v=R3eJW3PCYMs

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 by
Problem 19 		Followed by
Problem 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

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