Difference between revisions of "2000 AIME II Problems/Problem 3"

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== Solution ==
 
== Solution ==
There are <math>{38 \choose 2} = 703</math> ways we can draw a two cards from the deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in <math>9{4 \choose 2} = 54</math> ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in <math>1</math> way. Thus, the answer is <math>\frac{54+1}{703} = \frac{55}{703}</math>, and <math>m+n = \boxed{758}</math>.
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There are <math>{38 \choose 2} = 703</math> ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in <math>9{4 \choose 2} = 54</math> ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in <math>1</math> way. Thus, the answer is <math>\frac{54+1}{703} = \frac{55}{703}</math>, and <math>m+n = \boxed{758}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Probability Problems]]
 
[[Category:Intermediate Probability Problems]]
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{{MAA Notice}}

Revision as of 20:30, 4 July 2013

Problem

A deck of forty cards consists of four $1$'s, four $2$'s,..., and four $10$'s. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution

There are ${38 \choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\frac{54+1}{703} = \frac{55}{703}$, and $m+n = \boxed{758}$.

See also

2000 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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