2013 AIME I #6: Empty Boxes and Books

Question

Melinda has three empty boxes and $12$ 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 $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

I honestly have no idea how to even start this question. I thought about using stars and bars, but I have no idea how to apply it here.

score 2 · Answer 1 · answered May 31 '18 at 00:57

2

The easiest thing here is probably to solve the following equivalent problem: there are twelve numbered boxes in a row. Three are chosen at random. What is the probability that those three are numbers 1-3, three of numbers 4-7, or three of numbers 8-12?

answered May 31 '18 at 00:57

BallBoy

14,472

....Out of the box thinking.... :) – Dude156 May 31 '18 at 01:54
Incredible thought process. – DynamoBlaze May 31 '18 at 18:34

score 2 · Accepted Answer · answered May 31 '18 at 00:59

Consider ordering the $12$ books arbitrarily into one big stack, in one of $12!$ ways. Then, the top three books go into the small box, the next four into the middle box, and the last five into the big box.

There are $3!9!$ ways for the three math books to be on top. There are ${4\choose 3}3!9!$ ways for the three math books to be among the next four books. There are ${5\choose 3}3!9!$ ways for the three math books to be among the last five books.

Putting it all together, we get $$\frac{3!9!+{4\choose 3}3!9!+{5\choose 3}3!9!}{12!}$$

Now, simplify this, and sum the numerator and denominator.

That was a really cool way to think about this problem! Thanks a bunch! — Dude156, May 31 '18 at 01:53

2013 AIME I #6: Empty Boxes and Books

2 Answers2