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Difference between revisions of "2005 AMC 10B Problems/Problem 21"

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Forty slips are placed into a hat, each bearing a number <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>9</math>, or <math>10</math>, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let <math>p</math> be the probability that all four slips bear the same number. Let <math>q</math> be the probability that two of the slips bear a number <math>a</math> and the other two bear a number <math>b \neq a</math>. What is the value of <math>q/p</math>?
 
Forty slips are placed into a hat, each bearing a number <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>9</math>, or <math>10</math>, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let <math>p</math> be the probability that all four slips bear the same number. Let <math>q</math> be the probability that two of the slips bear a number <math>a</math> and the other two bear a number <math>b \neq a</math>. What is the value of <math>q/p</math>?
  
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<math>\mathrm{(A)} 162 \qquad \mathrm{(B)} 180 \qquad \mathrm{(C)} 324 \qquad \mathrm{(D)} 360 \qquad \mathrm{(E)} 720 </math>
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<math>\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720 </math>
  
 
== Solution 1 (where the order of drawing slips matters) ==
 
== Solution 1 (where the order of drawing slips matters) ==
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There are <math>{10 \choose 2} = 45</math> ways to determine which two numbers to pick for the second probability. There are <math>{4 \choose 2} = 6</math> ways to arrange the order which we draw the non-equal slips, and in each order there are <math>4 \times 3 \times 4 \times 3</math> ways to pick the slips, so <math>q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}</math>.  
 
There are <math>{10 \choose 2} = 45</math> ways to determine which two numbers to pick for the second probability. There are <math>{4 \choose 2} = 6</math> ways to arrange the order which we draw the non-equal slips, and in each order there are <math>4 \times 3 \times 4 \times 3</math> ways to pick the slips, so <math>q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}</math>.  
  
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Hence, the answer is <math>\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\ \mathbf{(A)}162}</math>.
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Hence, the answer is <math>\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\textbf{(A) }162}</math>.
  
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==Solution 2 (order does not matter)==
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==Solution 2 (where the order does not matter)==
  
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For probability <math>p</math>, there are <math>\binom{10}{1}=10</math> ways to choose the card you want to show up <math>4</math> times..
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For probability <math>p</math>, there are <math>\binom{10}{1}=10</math> ways to choose the number you want to show up <math>4</math> times.
  
 
Hence, the probability is <math>\frac{10}{\binom{40}{4}}</math>.
 
Hence, the probability is <math>\frac{10}{\binom{40}{4}}</math>.
  
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For probability <math>q</math>, there are <math>\binom{10}{2}=45</math> ways to choose the <math>2</math> numbers you want to show up twice. There are <math>\binom{4}{2}\cdot\binom{4}{2}</math> to pick which cards you want out of the <math>4</math> of each.
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For probability <math>q</math>, there are <math>\binom{10}{2}=45</math> ways to choose the <math>2</math> numbers you want to show up twice. There are <math>\binom{4}{2}\cdot\binom{4}{2}</math> ways to pick which slips you want out of the <math>4</math> of each.
  
 
Hence, the probability is <math>\frac{45\cdot6\cdot6}{\binom{40}{4}}</math>
 
Hence, the probability is <math>\frac{45\cdot6\cdot6}{\binom{40}{4}}</math>
  
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Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=162\implies \boxed{A}</math>.
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Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=\boxed{\textbf{(A) }162}</math>.
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/wopflrvUN2c?t=252
 +
 
 +
~ pi_is_3.14
 +
 
 +
== Video Solution ==
 +
https://www.youtube.com/watch?v=HVUV6NgH3wU  ~David
 +
 
 +
==See Also==
  
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== See Also ==
 
 
{{AMC10 box|year=2005|ab=B|num-b=20|num-a=22}}
 
{{AMC10 box|year=2005|ab=B|num-b=20|num-a=22}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:17, 17 September 2023

Problem

Forty slips are placed into a hat, each bearing a number $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, or $10$, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \neq a$. What is the value of $q/p$?

$\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720$

Solution 1 (where the order of drawing slips matters)

There are $10$ ways to determine which number to pick. There are $4!$ ways to then draw those four slips with that number, and $40 \cdot 39 \cdot 38 \cdot 37$ total ways to draw four slips. Thus $p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}$.

There are ${10 \choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 \choose 2} = 6$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 \times 3 \times 4 \times 3$ ways to pick the slips, so $q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}$.

Hence, the answer is $\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\textbf{(A) }162}$.

Solution 2 (where the order does not matter)

For probability $p$, there are $\binom{10}{1}=10$ ways to choose the number you want to show up $4$ times.

Hence, the probability is $\frac{10}{\binom{40}{4}}$.

For probability $q$, there are $\binom{10}{2}=45$ ways to choose the $2$ numbers you want to show up twice. There are $\binom{4}{2}\cdot\binom{4}{2}$ ways to pick which slips you want out of the $4$ of each.

Hence, the probability is $\frac{45\cdot6\cdot6}{\binom{40}{4}}$

Hence, $\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=\boxed{\textbf{(A) }162}$.

Video Solution by OmegaLearn

https://youtu.be/wopflrvUN2c?t=252

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=HVUV6NgH3wU ~David

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
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