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Difference between revisions of "1988 AHSME Problems/Problem 18"

(Created page with "==Problem== At the end of a professional bowling tournament, the top 5 bowlers have a playoff. First #5 bowls #4. The loser receives <math>5</math>th prize and the winner bowls...")
 
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==Solution==
 
==Solution==
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We have <math>2</math> choices for who wins the first game, and that uniquely determines <math>5^{\text{th}}</math> place. Then there are <math>2</math> choices for a next game and that uniquely determines <math>4^{\text{th}}</math> place, followed by <math>2</math> choices for the next game that uniquely determines <math>3^{\text{rd}}</math> place. Finally, there are <math>2</math> choices for the last game, which uniquely determines both <math>1^{\text{st}}</math> and <math>2^{\text{nd}}</math> places, since the winner is <math>1^{\text{st}}</math> and the loser is <math>2^{\text{nd}}</math>. Thus the number of possible orders is <math>2 \times 2 \times 2 \times 2 = 16</math>, which is <math>\boxed{\text{B}}</math>.
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== See also ==
 
== See also ==

Latest revision as of 14:03, 27 February 2018

Problem

At the end of a professional bowling tournament, the top 5 bowlers have a playoff. First #5 bowls #4. The loser receives $5$th prize and the winner bowls #3 in another game. The loser of this game receives $4$th prize and the winner bowls #2. The loser of this game receives $3$rd prize and the winner bowls #1. The winner of this game gets 1st prize and the loser gets 2nd prize. In how many orders can bowlers #1 through #5 receive the prizes?

$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 120\qquad \textbf{(E)}\ \text{none of these}$


Solution

We have $2$ choices for who wins the first game, and that uniquely determines $5^{\text{th}}$ place. Then there are $2$ choices for a next game and that uniquely determines $4^{\text{th}}$ place, followed by $2$ choices for the next game that uniquely determines $3^{\text{rd}}$ place. Finally, there are $2$ choices for the last game, which uniquely determines both $1^{\text{st}}$ and $2^{\text{nd}}$ places, since the winner is $1^{\text{st}}$ and the loser is $2^{\text{nd}}$. Thus the number of possible orders is $2 \times 2 \times 2 \times 2 = 16$, which is $\boxed{\text{B}}$.

See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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