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>? | ||
− | <math>\ | + | <math>\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720 </math> |
− | == Solution (where the order of drawing slips matters) == | + | == 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</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>. | 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{\ | + | 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 (order does not matter)== | + | ==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 | + | 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>. | ||
− | 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 | + | 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> | ||
− | Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3= | + | 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}} | {{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 19:17, 17 September 2023
Contents
Problem
Forty slips are placed into a hat, each bearing a number ,
,
,
,
,
,
,
,
, or
, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let
be the probability that all four slips bear the same number. Let
be the probability that two of the slips bear a number
and the other two bear a number
. What is the value of
?
Solution 1 (where the order of drawing slips matters)
There are ways to determine which number to pick. There are
ways to then draw those four slips with that number, and
total ways to draw four slips. Thus
.
There are ways to determine which two numbers to pick for the second probability. There are
ways to arrange the order which we draw the non-equal slips, and in each order there are
ways to pick the slips, so
.
Hence, the answer is .
Solution 2 (where the order does not matter)
For probability , there are
ways to choose the number you want to show up
times.
Hence, the probability is .
For probability , there are
ways to choose the
numbers you want to show up twice. There are
ways to pick which slips you want out of the
of each.
Hence, the probability is
Hence, .
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) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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