Difference between revisions of "2000 AIME II Problems/Problem 3"
| (7 intermediate revisions by 6 users not shown) | |||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| + | A deck of forty cards consists of four <math>1</math>'s, four <math>2</math>'s,..., and four <math>10</math>'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 <math>m/n</math> be the [[probability]] that two randomly selected cards also form a pair, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n.</math> | ||
== Solution == | == Solution == | ||
| + | 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 == | ||
| − | + | {{AIME box|year=2000|n=II|num-b=2|num-a=4}} | |
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
| + | [[Category:Intermediate Probability Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:30, 4 July 2013
Problem
A deck of forty cards consists of four 
's, four 
's,..., and four 
'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 
be the probability that two randomly selected cards also form a pair, where 
and 
are relatively prime positive integers. Find 
Solution
There are 
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 
ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in way. Thus, the answer is 
, and 
.
See also
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
