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Difference between revisions of "2024 AMC 12A Problems/Problem 16"

(Created page with "A set of <math>12</math> tokens ---- <math>3</math> red, <math>2</math> white, <math>1</math> blue, and <math>6</math> black ---- is to be distributed at random to <math>3</ma...")
 
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A set of <math>12</math> tokens ---- <math>3</math> red, <math>2</math> white, <math>1</math> blue, and <math>6</math> black ---- is to be distributed at random to <math>3</math> game players, <math>4</math> tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
+
==Problem==
 +
A set of <math>12</math> tokens — <math>3</math> red, <math>2</math> white, <math>1</math> blue, and <math>6</math> black — is to be distributed at random to <math>3</math> game players, <math>4</math> tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
 +
 
 
<math>
 
<math>
 
\textbf{(A) }387 \qquad
 
\textbf{(A) }387 \qquad
Line 8: Line 10:
 
</math>
 
</math>
  
−
==Solution 1 (12fact bash)==
+
==Solution 1A (Trivial/Easy solve)==
 +
We have <math>\binom{12}{4,4,4}</math> ways to handle the red/white/blue balls distribution on the denominator.
 +
Now we simply <math>\binom{6}{1}</math> <math>\binom{5}{2}</math> <math>3!</math> for the numerator in order to handle the black balls and distinguishable persons.
 +
The solution is therefore <math>\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}</math> or <math>4+385=\boxed{\textbf{(C) }389}.</math>
 +
 
 +
Remarks - Notice we let balls and persons be distinguishable to increase ease of calculations
 +
 
 +
~polya_mouse
 +
 
 +
==Solution 1B (12fact bash)==
  
 
We have <math>12!</math> total possible arrangements of <math>12</math> distinct tokens. If we imagine the first <math>4</math> tokens of our arrangement go to the first player, the next <math>4</math> go to the second, and the final <math>4</math> go to the third, then we can view this problem as counting the number of valid arrangements.  
 
We have <math>12!</math> total possible arrangements of <math>12</math> distinct tokens. If we imagine the first <math>4</math> tokens of our arrangement go to the first player, the next <math>4</math> go to the second, and the final <math>4</math> go to the third, then we can view this problem as counting the number of valid arrangements.  
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Firstly, the tokens are not all distinct, so we multiply by <math>3!</math>, <math>2!</math>, <math>1!</math>, and <math>6!</math> to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.  
 
Firstly, the tokens are not all distinct, so we multiply by <math>3!</math>, <math>2!</math>, <math>1!</math>, and <math>6!</math> to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.  
  
−
Letting <math>R</math> denote red, <math>W</math> denote white, <math>B</math> denote blue, and <math>L</math> denote black, then our arrangement must be something like <math>RRRLWWLLWLLL</math>. The three players are arbitrary, so we multiply by <math>3!</math>; then, the player who gets the reds has <math>\dbinom41=4</math> possible arrangements, the player who gets the whites has <math>\dbinom42=6</math> possibilities, and the player who gets the blacks has <math>\dbinom43=4</math> possibilities. Our total on top is thus <math>3!\cdot2!\cdot1!\cdot6!\cdot3!\cdot4\cdot6\cdot4</math>, and the denominator is <math>12!</math>. Firstly, we have the <math>6!</math> in the numerator cancel out part of the denominator; we thus have the following:  
+
Letting <math>R</math> denote red, <math>W</math> denote white, <math>B</math> denote blue, and <math>L</math> denote black, then our arrangement must be something like <math>RRRLWWLLBLLL</math>. The three players are arbitrary, so we multiply by <math>3!</math>; then, the player who gets the reds has <math>\dbinom41=4</math> possible arrangements, the player who gets the whites has <math>\dbinom42=6</math> possibilities, and the player who gets the blacks has <math>\dbinom43=4</math> possibilities. Our total on top is thus <math>3!\cdot2!\cdot1!\cdot6!\cdot3!\cdot4\cdot6\cdot4</math>, and the denominator is <math>12!</math>. Firstly, we have the <math>6!</math> in the numerator cancel out part of the denominator; we thus have the following:  
  
 
<cmath>\dfrac{3\cdot2\cdot2\cdot3\cdot2\cdot4\cdot6\cdot4}{12\cdot11\cdot10\cdot9\cdot8\cdot7}=\dfrac{2^83^3}{2^63^35\cdot7\cdot11}=\dfrac4{385}.</cmath>
 
<cmath>\dfrac{3\cdot2\cdot2\cdot3\cdot2\cdot4\cdot6\cdot4}{12\cdot11\cdot10\cdot9\cdot8\cdot7}=\dfrac{2^83^3}{2^63^35\cdot7\cdot11}=\dfrac4{385}.</cmath>
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~Technodoggo
 
~Technodoggo
 +
 +
==Solution 2==
 +
Assume all of them are distinct even though some have the same color,
 +
 +
Total possibility = <math>\mathrm{C}_{4}^{12}\mathrm{C}_{4}^{8}\mathrm{C}_{4}^{4}</math>
 +
(choosing 4 random token for each person)
 +
 +
Next, assume that all the token are already in 3 different groups (Note: 3! Ways to do so since 3 people)
 +
 +
We then distribute the 6 distinct black token into these 3 different groups (So 1,2,3 token for each group)
 +
 +
There are a total of <math>3!  \cdot  \mathrm{C}_{3}^{6}\mathrm{C}_{2}^{3}\mathrm{C}_{1}^{1}</math> ways in doing so
 +
 +
Thus the answer is <math>3! \cdot \frac{6!}{1!2!3!}/\frac{12!}{4!4!4!}=\frac{4}{385}</math>
 +
 +
So the answer is <math>\boxed{\textbf{(C) }389}</math>
 +
 +
~lptoggled
 +
 +
==Solution 3==
 +
We first assume there are designated red, white, and blue token players that will receive all of their respective one.
 +
 +
Consider each non-black token: The probability of the red player getting the first red token is <math>\frac{4}{12}</math>, because each player has 4 empty token "slots" for a total of 12. It follows that the probability of the player receiving all 3 red tokens is <math>\left(\frac{4}{12}\right)\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)</math>, the white token player is <math>\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)</math>, and the blue token player is <math>\frac{4}{7}</math>.
 +
 +
The combined probability is <math>\left(\frac{4}{12}\right)\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)\left(\frac{4}{7}\right)=\frac{2}{1155}</math>.
 +
 +
Finally, we multiply the probability by <math>3!=6</math> to remove our initial assumption to get <math>\frac{4}{385}</math>.
 +
 +
The requested sum is <math>4+385=\boxed{\textbf{(C) }389}</math>.
 +
 +
~SilverRush
 +
==Solution 4==
 +
Process start, <i><b>first</b></i> player get tokens.  The probability that he gets 3 red tokens and 1 black token is <math>\frac{3 \cdot 2 \cdot 1 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9} = \frac {1}{330}.</math>
 +
 +
There is  <math>\dbinom41 = 4</math> possible arrangements (RRRB,RRBR,RBRR,BRRR) and 3 possibilities who is the first, so the probability that <i><b>some</b></i> player gets 3 red tokens and 1 black token is <math>\frac {4 \cdot 3}{330} = \frac {2}{55}.</math>
 +
 +
After that <i><b>second</b></i> player get tokens. The probability that he gets 2 white tokens and 2 black tokens is <math>\frac{2 \cdot 1 \cdot 5 \cdot 4}{8 \cdot 7 \cdot 6 \cdot 5} = \frac {1}{42}.</math>
 +
 +
There is <math>\dbinom42 = 6</math> possible arrangements (WWBB,WBWB, WBBW, BBWW, BWBW, BWWB) and 2 possibilities who is the second, so the probability that <i><b>some</b></i> player gets 2 white tokens and 2 black tokens is <math>\frac {6 \cdot 2}{42} = \frac {2}{7}.</math>
 +
 +
The third player gets last tokens - 1 blue and 3 black tokens.
 +
 +
The desired probability is <math>\frac {2 \cdot 2}{55 \cdot 7} = \frac {4}{385} \implies 4+385=\boxed{\textbf{(C) }389}</math>.
 +
 +
To check the result suppose that first and some) player gets 1 blue and 3 black tokens. The probability is <math>\frac{1 \cdot 6 \cdot 5 \cdot 4}{12 \cdot 11 \cdot 10 \cdot 9} \cdot 4 \cdot 3 = \frac {4}{33}.</math>
 +
 +
The probability that second (and some) player gets 3 red tokens and 1 black token is <math>\frac{3 \cdot 2 \cdot 1 \cdot 3}{8 \cdot 7 \cdot 6 \cdot 5} \cdot 4 \cdot 2 = \frac {3}{35}.</math>
 +
 +
The desired probability is <math>\frac {4 \cdot 3}{33 \cdot 35} = \frac {4}{385}.</math>
 +
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=ynMtiJuLCNI
 +
 +
==See also==
 +
{{AMC12 box|year=2024|ab=A|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Latest revision as of 15:46, 17 November 2024

Problem

A set of $12$ tokens — $3$ red, $2$ white, $1$ blue, and $6$ black — is to be distributed at random to $3$ game players, $4$ tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A) }387 \qquad \textbf{(B) }388 \qquad \textbf{(C) }389 \qquad \textbf{(D) }390 \qquad \textbf{(E) }391 \qquad$

Solution 1A (Trivial/Easy solve)

We have $\binom{12}{4,4,4}$ ways to handle the red/white/blue balls distribution on the denominator. Now we simply $\binom{6}{1}$ $\binom{5}{2}$ $3!$ for the numerator in order to handle the black balls and distinguishable persons. The solution is therefore $\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}$ or $4+385=\boxed{\textbf{(C) }389}.$

Remarks - Notice we let balls and persons be distinguishable to increase ease of calculations

~polya_mouse

Solution 1B (12fact bash)

We have $12!$ total possible arrangements of $12$ distinct tokens. If we imagine the first $4$ tokens of our arrangement go to the first player, the next $4$ go to the second, and the final $4$ go to the third, then we can view this problem as counting the number of valid arrangements.

Firstly, the tokens are not all distinct, so we multiply by $3!$, $2!$, $1!$, and $6!$ to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.

Letting $R$ denote red, $W$ denote white, $B$ denote blue, and $L$ denote black, then our arrangement must be something like $RRRLWWLLBLLL$. The three players are arbitrary, so we multiply by $3!$; then, the player who gets the reds has $\dbinom41=4$ possible arrangements, the player who gets the whites has $\dbinom42=6$ possibilities, and the player who gets the blacks has $\dbinom43=4$ possibilities. Our total on top is thus $3!\cdot2!\cdot1!\cdot6!\cdot3!\cdot4\cdot6\cdot4$, and the denominator is $12!$. Firstly, we have the $6!$ in the numerator cancel out part of the denominator; we thus have the following:

\[\dfrac{3\cdot2\cdot2\cdot3\cdot2\cdot4\cdot6\cdot4}{12\cdot11\cdot10\cdot9\cdot8\cdot7}=\dfrac{2^83^3}{2^63^35\cdot7\cdot11}=\dfrac4{385}.\]

Our answer is $4+385=\boxed{\textbf{(C) }389}.$

~Technodoggo

Solution 2

Assume all of them are distinct even though some have the same color,

Total possibility = $\mathrm{C}_{4}^{12}\mathrm{C}_{4}^{8}\mathrm{C}_{4}^{4}$ (choosing 4 random token for each person)

Next, assume that all the token are already in 3 different groups (Note: 3! Ways to do so since 3 people)

We then distribute the 6 distinct black token into these 3 different groups (So 1,2,3 token for each group)

There are a total of $3! \cdot \mathrm{C}_{3}^{6}\mathrm{C}_{2}^{3}\mathrm{C}_{1}^{1}$ ways in doing so

Thus the answer is $3! \cdot \frac{6!}{1!2!3!}/\frac{12!}{4!4!4!}=\frac{4}{385}$

So the answer is $\boxed{\textbf{(C) }389}$

~lptoggled

Solution 3

We first assume there are designated red, white, and blue token players that will receive all of their respective one.

Consider each non-black token: The probability of the red player getting the first red token is $\frac{4}{12}$, because each player has 4 empty token "slots" for a total of 12. It follows that the probability of the player receiving all 3 red tokens is $\left(\frac{4}{12}\right)\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)$, the white token player is $\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)$, and the blue token player is $\frac{4}{7}$.

The combined probability is $\left(\frac{4}{12}\right)\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)\left(\frac{4}{7}\right)=\frac{2}{1155}$.

Finally, we multiply the probability by $3!=6$ to remove our initial assumption to get $\frac{4}{385}$.

The requested sum is $4+385=\boxed{\textbf{(C) }389}$.

~SilverRush

Solution 4

Process start, first player get tokens. The probability that he gets 3 red tokens and 1 black token is $\frac{3 \cdot 2 \cdot 1 \cdot 6}{12 \cdot 11 \cdot 10 \cdot 9} = \frac {1}{330}.$

There is $\dbinom41 = 4$ possible arrangements (RRRB,RRBR,RBRR,BRRR) and 3 possibilities who is the first, so the probability that some player gets 3 red tokens and 1 black token is $\frac {4 \cdot 3}{330} = \frac {2}{55}.$

After that second player get tokens. The probability that he gets 2 white tokens and 2 black tokens is $\frac{2 \cdot 1 \cdot 5 \cdot 4}{8 \cdot 7 \cdot 6 \cdot 5} = \frac {1}{42}.$

There is $\dbinom42 = 6$ possible arrangements (WWBB,WBWB, WBBW, BBWW, BWBW, BWWB) and 2 possibilities who is the second, so the probability that some player gets 2 white tokens and 2 black tokens is $\frac {6 \cdot 2}{42} = \frac {2}{7}.$

The third player gets last tokens - 1 blue and 3 black tokens.

The desired probability is $\frac {2 \cdot 2}{55 \cdot 7} = \frac {4}{385} \implies 4+385=\boxed{\textbf{(C) }389}$.

To check the result suppose that first and some) player gets 1 blue and 3 black tokens. The probability is $\frac{1 \cdot 6 \cdot 5 \cdot 4}{12 \cdot 11 \cdot 10 \cdot 9} \cdot 4 \cdot 3 = \frac {4}{33}.$

The probability that second (and some) player gets 3 red tokens and 1 black token is $\frac{3 \cdot 2 \cdot 1 \cdot 3}{8 \cdot 7 \cdot 6 \cdot 5} \cdot 4 \cdot 2 = \frac {3}{35}.$

The desired probability is $\frac {4 \cdot 3}{33 \cdot 35} = \frac {4}{385}.$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution 1 by SpreadTheMathLove

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

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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All AMC 12 Problems and Solutions

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