AIME 2008 I · 第 9 题

Solution

Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:

$$\begin{aligned}3a + 4b + 6c &= 41\\ a + b + c &= 10\end{aligned}$$ Subtracting 3 times the second from the first gives $b + 3c = 11$, or $(b,c) = (2,3),(5,2),(8,1),(11,0)$. The last doesn't work, obviously. This gives the three solutions $(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$. In terms of choosing which goes where, the first two solutions are analogous.

For $(5,2,3),(3,5,2)$, we see that there are $2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7$ ways to stack the crates. For $(1,8,1)$, there are $\dfrac{10!}{8!1!1!} = 90$. Also, there are $3^{10}$ total ways to stack the crates to any height.

Thus, our probability is $\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}$. Our answer is the numerator, $\boxed{190}$.

1 Min Solution

It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that $3$ and $6$ are both divisible by $3$, so the number of $4$-crates must be congruent to $41\bmod{3}$, which is also congruent to $2\bmod{3}$. Our solutions for the number of $4$-crates will repeat mod $3$, so if $x$ is a solution, so is $x+3$. By inspection, we have that $2$ is solution, and so are $5$ and $8$. Each solution splits into its own case.We must solve the equation $41-4z=6x+3y$, simultaneously with $x+y=10-z$. Note that we already know the possible values of $z$. Solving these (it's AIME $9$, you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution sets $\{8,1,1\},\{5,2,3\},$ and $\{2,3,5\}$. We can count the number of possible arrangements for each solution by taking $\dbinom{10}{z}$ and then multiplying by $\dbinom{10-z}{x}$ (the solution sets, for the sake of consistency, are in the form $z,x,y$). Summing the results for all the solutions gives us $5130$. Finally, to calculate the probability we must determine our denominator. Since we have $3$ ways to arrange each block, our denominator is $3^{10}$. $\frac{5130}{3^{10}}=\frac{190}{3^7}$. The answer is $m=\boxed{190}$.

Solution 2 (a more direct approach)

Let's make two observations. We are trying to find the number of ways we can add $3\text{s}, 4\text{s}$, and $6\text{s}$ to get $41$, and the total number of (non-distinct) sums possible is $3^{10}$. Then we just use casework to easily and directly solve for the number of ways to get $41$. To begin, the minimum sum is produced with $10$ threes, so WLOG we can solve for the number of ways to get $11$ with $0\text{s}, 1\text{s}$, and $3\text{s}$.

Case I: $0$ zeroes, $0$ threes, $11$ ones Impossible, because there are only ten available spots.

Case II: $1$ zero, $1$ three, $8$ ones This is just $\frac{10!}{8!}$, so there are $90$ possibilities.

Case III: $3$ zeroes, $2$ threes, $5$ ones This is just $\frac{10!}{3!2!5!}$. This gives $2520$ possibilities.

Case IV: 5 zeroes, 3 threes, and 2 ones. This is the same as case $3$, so also $2520$ possibilities.

$90+2520+2520=5130$

$5130$ has three powers of $3$, so $5130$ divided by $27$ is $\boxed{190}$.

-jackshi2006

Solution 3 (Generating Functions)

Note we are placing 10 crates where each "height" is 3, 4, 6 and we want all the heights to sum to 41. We can model this as the generating function

$$\left(x^3+x^4+x^6\right)^{10}$$ where we want the coefficient of $x^{41}.$ First off, factor this to get

$${x^{30} \left(1 + x + x^3\right)^{10}}$$ and then see that we want the coefficient of $x^{11}$ in $\left(1+x+x^3\right)^{10}.$ From multinomial theorem, this expansion is

$$\sum_{a+b+c=10}\binom{10}{a,b,c}1^ax^bx^{3c}$$ If we want the coefficient of $x^{11}$ then we need $b + 3c = 11.$ with $b + c \le 10$(from the multinomial expansion). This has the solutions $(b, c) = \{8, 1\}, \{5, 2\}, \{2, 3\}.$ Note that the denominator of the answer is just $3^{10}$ since there are 3 ways to orientate every crate and there are 10 creates. Thus, our answer is

$$\frac{\binom{10}{8,1,1} + \binom{10}{5,2,3} + \binom{10}{2,3,5}}{3^{10}} = \frac{190}{3^7} \rightarrow \boxed{190}$$

Video Solution (Very fast)

https://www.youtube.com/watch?v=qeSY_ISSX6M&t=33s

~fidgetboss_4000