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

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

@@ Line 1: / Line 1: @@
 == Problem ==
-== Solution ==
+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>?
-== See Also ==
-*[[2005 AMC 10B Problems]]
+<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) ==
+There are <math>10</math> ways to determine which number to pick. There are <math>4!</math> ways to then draw those four slips with that number, and <math>40 \cdot 39 \cdot 38 \cdot 37</math> total ways to draw four slips. Thus <math>p = \frac{10\cdot 4!}{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>.
+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>.
+==Solution 2 (where the order does not matter)==
+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>.
+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, <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==
+{{AMC10 box|year=2005|ab=B|num-b=20|num-a=22}}
+[[Category:Introductory Combinatorics Problems]]
+{{MAA Notice}}

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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