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 $m+n$.

The total ways the textbooks can be arranged in the 3 boxes is $12\textbf{C}3\cdot 9\textbf{C}4$, which is equivalent to $\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6}{144}=12\cdot11\cdot10\cdot7\cdot3$. If all of the math textbooks are put into the box that can hold $3$ textbooks, there are $9!/(4!\cdot 5!)=9\textbf{C}4$ 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 $9$ ways to choose the other book in that box, times $8\textbf{C}3$ 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 $9\textbf{C}2$ ways to choose the other 2 textbooks in that box, times $7\textbf{C}3$ ways to arrange the other 7 textbooks. $9\textbf{C}4=9\cdot7\cdot2=126$, $9\cdot 8\textbf{C}3=9\cdot8\cdot7=504$, and $9\textbf{C}2\cdot 7\textbf{C}3=9\cdot7\cdot5\cdot4=1260$, so the total number of ways the math textbooks can all be placed into the same box is $126+504+1260=1890$. So, the probability of this occurring is $\frac{(9\cdot7)(2+8+(4\cdot5))}{12\cdot11\cdot10\cdot7\cdot3}=\frac{1890}{27720}$. If the numerator and denominator are both divided by $9\cdot7$, we have $\frac{(2+8+(4\cdot5))}{4\cdot11\cdot10}=\frac{30}{440}$. Simplifying the numerator yields $\frac{30}{10\cdot4\cdot11}$, and dividing both numerator and denominator by $10$ results in $\frac{3}{44}$. This fraction cannot be simplified any further, so $m=3$ and $n=44$. Therefore, $m+n=3+44=\boxed{47}$.
47
Level 5
Counting & Probability
Problem: 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 $m+n$.

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that all three mathematics textbooks end up in the same box, we first determine the total number of ways to distribute the 12 textbooks into the three boxes. The boxes have capacities of 3, 4, and 5 textbooks respectively. 

We can denote the boxes as Box A (3 textbooks), Box B (4 textbooks), and Box C (5 textbooks). The total number of ways to choose which textbooks go into each box can be calculated using combinations. 

1. Choose 3 textbooks for Box A from the 12 available. This can be done in C(12, 3) ways.
2. After placing 3 textbooks in Box A, 9 textbooks remain. Choose 4 textbooks for Box B from these 9. This can be done in C(9, 4) ways.
3. The remaining 5 textbooks will automatically go into Box C. 

Thus, the total number of ways to distribute the textbooks is given by:
Total ways = C(12, 3) * C(9, 4).

Next, we calculate the number of favorable outcomes where all three mathematics textbooks are in the same box. We consider three cases based on which box contains all three mathematics textbooks:

Case 1: All mathematics textbooks in Box A (3 textbooks). This is only possible if Box A contains exactly the 3 mathematics textbooks. The remaining 9 textbooks can be distributed into Box B and Box C. The number of ways to choose 0 textbooks for Box A (since it is full) is 1. The remaining 9 textbooks can be distributed into Box B and Box C in C(9, 4) ways.

Case 2: All mathematics textbooks in Box B (4 textbooks). We need to choose 1 additional textbook from the remaining 9 textbooks to go into Box B. The number of ways to choose 1 textbook is C(9, 1). The remaining 8 textbooks can be distributed into Box A and Box C. The number of ways to choose 3 textbooks for Box A from these 8 is C(8, 3).

Case 3: All mathematics textbooks in Box C (5 textbooks). We need to choose 2 additional textbooks from the remaining 9 textbooks to go into Box C. The number of ways to choose 2 textbooks is C(9, 2). The remaining 7 textbooks can be distributed into Box A and Box B. The number of ways to choose 3 textbooks for Box A from these 7 is C(7, 3).

Now we can summarize the favorable outcomes:
- For Case 1: C(9, 4)
- For Case 2: C(9, 1) * C(8, 3)
- For Case 3: C(9, 2) * C(7, 3)

The total number of favorable outcomes is:
Favorable outcomes = C(9, 4) + C(9, 1) * C(8, 3) + C(9, 2) * C(7, 3).

Finally, the probability that all three mathematics textbooks end up in the same box is given by the ratio of favorable outcomes to total outcomes:
Probability = Favorable outcomes / Total ways.

After calculating the values of the combinations and simplifying the fraction, we find that the probability can be expressed as m/n, where m and n are relatively prime integers. The final answer, m+n, is computed to be 47.<|END_OF_LATENT|> The final answer is $47$.