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

 
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== Problem ==
 
== Problem ==
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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>?
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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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== Solution ==
 
== Solution ==
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There are <math>10</math> ways to determine which number to pick. There are <math>4!</math> way 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{4!}{40 \cdot 39 \cdot 38 \cdot 37}</math>.
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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}</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 \times 39 \times 38 \times 37}</math>.
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Hence, the answer is <math>\frac{q}{p} = \frac{2^5 \cdot 3^4 \cdot 5}{4! \times 10} = 162\ \mathbf{(A)}</math>.
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== See Also ==
 
== See Also ==
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*[[2005 AMC 10B Problems]]
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{{AMC10 box|year=2005|ab=B|num-b=20|num-a=22}}
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[[Category:Introductory Combinatorics Problems]]

Revision as of 16:29, 7 February 2009

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 $q/p$?

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

Solution

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

There are ${10 \choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 \choose 2}$ 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 $q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \times 39 \times 38 \times 37}$.

Hence, the answer is $\frac{q}{p} = \frac{2^5 \cdot 3^4 \cdot 5}{4! \times 10} = 162\ \mathbf{(A)}$.

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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