Difference between revisions of "2005 AMC 10B Problems/Problem 21"
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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> | <math>\mathrm{(A)} 162 \qquad \mathrm{(B)} 180 \qquad \mathrm{(C)} 324 \qquad \mathrm{(D)} 360 \qquad \mathrm{(E)} 720 </math> | ||
| − | == Solution == | + | == Solution (where the order of drawing slips matters) == |
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{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> 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{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}</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>. | 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>. | ||
| − | Hence, the answer is <math>\frac{q}{p} = \frac{2^5 \cdot 3^4 \cdot 5}{10\cdot 4!} = | + | Hence, the answer is <math>\frac{q}{p} = \frac{2^5 \cdot 3^4 \cdot 5}{10\cdot 4!} = \boxed{\ \mathbf{(A)}162}</math>. |
== See Also == | == See Also == | ||
Revision as of 21:37, 26 November 2013
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 (where the order of drawing slips matters)
There are ways to determine which number to pick. There are 
way 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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
