Difference between revisions of "2021 April MIMC 10 Problems/Problem 13"
Cellsecret (talk | contribs) (Created page with "Given that Giant want to put <math>12</math> green identical balls into <math>3</math> different boxes such that each box contains at least two balls, and that no box can cont...") |
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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== | ||
| − | + | 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 
green identical balls into 
different boxes such that each box contains at least two balls, and that no box can contain 
or more balls. Find the number of ways that Giant can accomplish this.
Solution
We can use stars and bars to calculate the total amount of ways without thinking about the condition "no box can contain or more balls". Giving each box 
ball in advance to meet "each box contains at least two balls" requirement since each box will get at least extra, There are 
or 
ways to accomplish this. We need to subtract the situations where a box contains or more balls. There are two situations: 
in each box or 
in each box. In the first situation, since two boxes have the same amount, it will have 
ways of accomplishing it, and the second situation will have 
ways of accomplishing it. Therefore, the total amount of ways that meets the requirement is 
.
