2002 Indonesia MO Problems/Problem 2

Problem

Five regular dices are thrown, one at each time, then the product of the $5$ numbers shown are calculated. Which probability is bigger; the product is $180$ or the product is $144$ ?

Solution

Let $a$ be the roll of the first dice, $b$ be the roll of the second dice, $c$ be the roll of the third dice, $d$ be the roll of the fourth dice, and $e$ be the roll of the fifth dice. To calculate which probability is bigger, find the number of ways to roll dice that result in the two wanted values. Note that the prime factorization of $180$ is $2^2 \cdot 3^2 \cdot 5$ and the prime factorization of $144$ is $2^4 \cdot 3^2$ .

If the product of the five dices is , then , where . To find the number of ways, create casework based on the number of ones.
- For the case of no $1$ , the only way that works is $2,2,3,3,5$ , for a total of $\tfrac{5!}{2!2!} = 30$ possibilities.
- For the case of one $1$ , the two ways that work are $1,4,3,3,5$ and $1,6,2,3,5$ , for a total of $\tfrac{5!}{2!} + 5! = 60+120 = 180$ possibilities.
- For the case of two $1$ , the only way that works is $1,1,6,6,5$ , for a total of $\tfrac{5!}{2!2!} = 30$ possibilities.
If the product of the five dices is , then , where . To find the number of ways, create casework based on the number of ones.
- For the case of no $1$ , the two ways that work are $2,2,2,3,6$ and $4,2,2,3,3$ , for a total of $\tfrac{5!}{3!} + \tfrac{5!}{2!2!} = 20+30 = 50$ possibilities.
- For the case of one $1$ , the three ways that work are $1,4,4,3,3$ and $1,4,2,6,3$ and $1,2,2,6,6$ , for a total of $\tfrac{5!}{2!2!} + 5! + \tfrac{5!}{2!2!} = 30+120+30 = 180$ possibilities.
- For the case of two $1$ , the only way that works is $1,1,4,4,6$ , for a total of $\tfrac{5!}{2!2!} = 30$ possibilities.