In how many ways, Miyad can do the Trick? The numbers 1 to 6 are written on six balls. Miyad took four balls from there such that if he multiplies the numbers of the balls and then say it to Mun, then Mun can’t say the number of the balls surely. In how many ways, Miyad can do this?


This is a question of Math Olympiad and I'm slightly confused in line 2. Can someone please help?
The prime $5$ occurs in only one of the numbers, so Mun can tell whether it&rsquo;s one of the $4$ balls, and if it is, divide it out of the product. Thus, we can split the problem into two cases: Choose $4$ numbers without $5$ such that their product is not unique, or choose $3$ numbers in addition to $5$ such that their product is not unique.

The first case is impossible, since only one of the other $5$ numbers is omitted, so Mun can find that number by dividing the product of all $5$ other numbers by the given product.

For the second case, if we choose two different triples out of $5$ numbers, they have at least one number in common. They can&rsquo;t have the same product if they differ only in one number; so they must have exactly one number in common. The only two pairs of numbers up to $6$ that have the same product are $2\cdot3=1\cdot6$ and $2\cdot6=3\cdot4$.

Thus, there are exactly four ways in which Miyad can choose the numbers such that their product doesn&rsquo;t uniquely identify them, namely $\{1,4,5,6\}$ and $\{2,3,4,5\}$ (for $2\cdot3=1\cdot6$) and $\{1,2,5,6\}$ and $\{1,3,4,5\}$ (for $2\cdot6=3\cdot4$).
In general there are $\displaystyle{\binom{6}{4}=15}$ ways for Miyad to choose the balls.

For instance $\{1,2,3,4\}$ giving product $24$, or $\{2,3,5,6\}$, giving product 180.

If the product is unique to one selection, the balls can be identified with certainty, if the same product occurs with more than one selection, they can't be identified with certainty.