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Difference between revisions of "2021 April MIMC 10 Problems/Problem 13"

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<math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~19 </math>
 
<math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~19 </math>
 
==Solution==
 
==Solution==
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We can use stars and bars to calculate the total amount of ways without thinking about the condition "no box can contain <math>7</math> or more balls". Giving each box <math>1</math> ball in advance to meet "each box contains at least two balls" requirement since each box will get at least <math>1</math> extra, There are <math>8\choose2</math> or <math>28</math> ways to accomplish this. We need to subtract the situations where a box contains <math>7</math> or more balls. There are two situations: <math>2,2,8</math> in each box or <math>2,3,7</math> in each box. In the first situation, since two boxes have the same amount, it will have <math>\frac{3!}{2}=3</math> ways of accomplishing it, and the second situation will have <math>3!=6</math> ways of accomplishing it. Therefore, the total amount of ways that meets the requirement is <math>28-3-6=\fbox{\textbf{(E)} 19}</math>.
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We can use stars and bars to calculate the total amount of ways without thinking about the condition "no box can contain <math>7</math> or more balls". Giving each box <math>1</math> ball in advance to meet "each box contains at least two balls" requirement since each box will get at least <math>1</math> extra, There are <math>{8\choose2}</math> or <math>28</math> ways to accomplish this. We need to subtract the situations where a box contains <math>7</math> or more balls. There are two situations: <math>2,2,8</math> in each box or <math>2,3,7</math> in each box. In the first situation, since two boxes have the same amount, it will have <math>\frac{3!}{2}=3</math> ways of accomplishing it, and the second situation will have <math>3!=6</math> ways of accomplishing it. Therefore, the total amount of ways that meets the requirement is <math>28-3-6=\fbox{\textbf{(E)} 19}</math>.

Latest revision as of 12:43, 26 April 2021

Given that Giant want to put $12$ green identical balls into $3$ different boxes such that each box contains at least two balls, and that no box can contain $7$ or more balls. Find the number of ways that Giant can accomplish this.

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~19$

Solution

We can use stars and bars to calculate the total amount of ways without thinking about the condition "no box can contain $7$ or more balls". Giving each box $1$ ball in advance to meet "each box contains at least two balls" requirement since each box will get at least $1$ extra, There are ${8\choose2}$ or $28$ ways to accomplish this. We need to subtract the situations where a box contains $7$ or more balls. There are two situations: $2,2,8$ in each box or $2,3,7$ in each box. In the first situation, since two boxes have the same amount, it will have $\frac{3!}{2}=3$ ways of accomplishing it, and the second situation will have $3!=6$ ways of accomplishing it. Therefore, the total amount of ways that meets the requirement is $28-3-6=\fbox{\textbf{(E)} 19}$.

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