Question

Ten identical crates each have dimensions $3$ ft $\times$ $4$ ft $\times$ $6$ ft. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac{m}{n}$ be the probability that the stack of crates is exactly $41$ ft tall, where $m$ and $n$ are relatively prime positive integers. Find $m.$

答案

Solution:

Each crate independently contributes height $3,$ $4,$ or $6,$ each with probability $\frac{1}{3},$ so there are $3^{10}$ equally likely stacks. If $x,$ $y,$ $z$ crates have heights $3,$ $4,$ $6,$ then $x + y + z = 10$ and $3x + 4y + 6z = 41;$ subtracting three times the first equation gives $y + 3z = 11,$ so $(x, y, z) = (1, 8, 1), \quad (3, 5, 2), \quad (5, 2, 3).$

These can be ordered in $\frac{10!}{1!\,8!\,1!} = 90,$ $\frac{10!}{3!\,5!\,2!} = 2520,$ and $\frac{10!}{5!\,2!\,3!} = 2520$ ways, for $5130$ stacks in all. The probability is $\frac{5130}{3^{10}} = \frac{190}{3^7},$ which is in lowest terms since $190 = 2 \cdot 5 \cdot 19.$ Thus $m = 190.$

2008 AIME I Problem 9

Solution:

Problem 9 in Other Years