Difference between revisions of "2013 AIME I Problems/Problem 6"
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The math books are placed into the largest box. This can be done in <math>\binom{5}{3}</math> ways. | The math books are placed into the largest box. This can be done in <math>\binom{5}{3}</math> ways. | ||
| − | So, the total ways the condition can be satisfied is <math>\binom{3}{3} + \binom{4}{3} + \binom{5}{3}</math>. This can be simplified to <math>\binom{6}{4} = \binom{6}{2}</math> by the Hockey Stick Identity. Therefore, the desired probability is <math>\ | + | So, the total ways the condition can be satisfied is <math>\binom{3}{3} + \binom{4}{3} + \binom{5}{3}</math>. This can be simplified to <math>\binom{6}{4} = \binom{6}{2}</math> by the Hockey Stick Identity. Therefore, the desired probability is <math>\dfrac{\binom{6}{2} }{\binom{12}{3}}</math> = <math>\dfrac{3}{44}</math>, and <math>m+n=3+44=\boxed{047}</math>. |
== See also == | == See also == | ||
Revision as of 22:29, 27 February 2017
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 
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 
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 
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 
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. 
