Difference between revisions of "2013 AIME I Problems/Problem 6"
Mathgeek2006 (talk | contribs) m (→See also) |
|||
Line 19: | Line 19: | ||
== See also == | == See also == | ||
− | {{AIME box|year=2013|n=I| | + | {{AIME box|year=2013|n=I|num-b=5|num-a=7}} |
{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 17:53, 13 March 2015
Problem 6
Melinda has three empty boxes and textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as , where and are relatively prime positive integers. Find .
Solution One
The total ways the textbooks can be arranged in the 3 boxes is , which is equivalent to . If all of the math textbooks are put into the box that can hold 3 textbooks, there are ways for the other textbooks to be arranged. If all of the math textbooks are put into the box that can hold 4 textbooks, there are ways to choose the other book in that box, times ways for the other books to be arranged. If all of the math textbooks are put into the box with the capability of holding 5 textbooks, there are ways to choose the other 2 textbooks in that box, times ways to arrange the other 7 textbooks. , , and , so the total number of ways the math textbooks can all be placed into the same box is . So, the probability of this occurring is . If the numerator and denominator are both divided by , we have . Simplifying the numerator yields , and dividing both numerator and denominator by 10 results in . This fraction cannot be simplified any further, so and . Therefore, .
Solution Two
Consider the books as either math or not-math where books in each category are indistiguishable from one another. Then, there are total distinguishable ways to pack the books. Now, in order to determine the desired propability, we must find the total number of ways the condition that all math books are in the same box can be satisfied. We proceed with casework for each box:
Case 1: The math books are placed into the smallest box. This can be done in ways.
Case 2: The math books are placed into the middle box. This can be done in ways.
Case 3: The math books are placed into the largest box. This can be done in ways.
So, the total ways the condition can be satisfied is . This can be simplified to by the Hockey Stick Identity. Therefore, the desired probability is = , and .
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.