Number of possible products

The number of products consisting of $3$ elements with repetition from the $8$ element set \begin{align*} \{2,3,4,5,6,7,8,9\}=\{2,2^2,2^3,2\cdot 3, 3,3^2,5,7\} \end{align*} is \begin{align*} \binom{8+3-1}{3}=\binom{10}{3}=\color{blue}{120} \end{align*}

Here the strategy is to count the products which occur more than once manually and then subtract this number from $120$.

• We can see at a glance that products having two elements from $\{5,7\}$ cannot have two different representations. Therefore candidates have either a factor $5$ or $7$ or none of them.

We need some kind of systematic way to find all possible products with multiple representations. In order to do so we arrange all candidates according to two criterias:

• The product contains either a factor $5$ or $7$ or none of them

• We sort candidates by increasing powers of $2$

Products containing $5$ or $7$:

We only need to list two factors, since the third one is either $5$ or $7$. We obtain \begin{align*} \text{factor }2\ :&\qquad 2\cdot 3^2=(2\cdot 3)\cdot 3&(90,126)\\ \text{factor }2^2:&\qquad 2^2\cdot3=2\cdot(2\cdot 3)&(60,84)\\ &\qquad 2^2\cdot 3^2=(2\cdot 3)(2\cdot 3)&(180,252)\\ \text{factor }2^3:&\qquad 2^3\cdot 3=2^2\cdot (2\cdot 3)&(120, 168)\\ \text{factor }2^4:&\qquad 2\cdot 2^3=2^2\cdot 2^2&(80,112)\\ \end{align*}

We see there are $5$ possibilities to write a product with two different representations. Since the third factor is either $5$ or $7$ we have a total of $\color{blue}{10}$ possibilities. The bracketed values on the right of each line give the number with factor $5$ resp. $7$.

Products containing neither $5$ nor $7$:

\begin{align*} \text{factor }2\ :&\qquad 2\cdot 3\cdot 3^2=(2\cdot 3)\cdot 3\cdot 3=54\\ &\qquad 2\cdot 3^2\cdot 3^2=(2\cdot 3)\cdot 3\cdot 3^2=162\\ &\qquad \rightarrow 2\text{ multiples}\\ \text{factor }2^2:&\qquad 2\cdot 2\cdot 3^2=2^2\cdot 3\cdot 3=2\cdot (2\cdot 3)\cdot 3=36\\ &\qquad 2^2\cdot 3\cdot 3^2=2\cdot (2\cdot 3)\cdot 3^2=(2\cdot 3)(2\cdot 3)\cdot 3=108\\ &\qquad 2^2\cdot 3^2\cdot 3^2=(2\cdot 3)(2\cdot 3)\cdot 3^2=324\\ &\qquad \rightarrow 5\text{ multiples}\\ \text{factor }2^3:&\qquad 2\cdot 2^2\cdot 3=2\cdot 2\cdot (2\cdot 3)=24\\ &\qquad 2\cdot 2^2 \cdot 3^2=2^2\cdot(2\cdot3)\cdot 3=2^3\cdot 3\cdot 3=2\cdot(2\cdot3)(2\cdot 3)=72\\ &\qquad 2^3\cdot 3\cdot 3^2=2^2\cdot (2\cdot 3)\cdot 3^2=(2\cdot 3)(2\cdot 3)(2\cdot 3)=216\\ &\qquad \rightarrow 6\text{ multiples}\\ \text{factor }2^4:&\qquad 2\cdot2^3\cdot 3=2^2\cdot2^2\cdot 3=2\cdot 2^2\cdot (2\cdot 3)=48\\ &\qquad 2\cdot 2^3\cdot 3^2=2^2\cdot2^2\cdot 3^2=2^3\cdot(2\cdot3)\cdot3=2^2\cdot (2\cdot 3)(2\cdot 3)=144\\ &\qquad \rightarrow 5\text{ multiples}\\ \text{factor }2^5:&\qquad 2\cdot 2\cdot2^3=2\cdot2^2\cdot2^2=32\\ &\qquad 2^2\cdot 2^3\cdot 3=2\cdot 2^3\cdot (2\cdot 3)=2^2\cdot 2^2\cdot (2\cdot 3)=96\\ &\qquad 2^2\cdot 2^3\cdot 3^2=2^3\cdot (2\cdot 3)(2\cdot 3)=288\\ &\qquad \rightarrow 4\text{ multiples}\\ \text{factor }2^6:&\qquad 2\cdot 2^2\cdot 2^3=2^2\cdot 2^2\cdot 2^2=64\\ &\qquad 2^3\cdot 2^3\cdot 3=2^2\cdot 2^3\cdot (2\cdot 3)=192\\ &\qquad \rightarrow 2\text{ multiples}\\ \text{factor }2^7:&\qquad 2\cdot 2^3\cdot 2^3=2^2\cdot 2^2\cdot 2^3=128\\ &\qquad \rightarrow 1\text{ multiple}\\ \end{align*} giving a total of $2+5+6+5+4+2+1=\color{blue}{25}$ multiple product representations.

We conclude the number of different products is $$120-10-25=\color{blue}{85}.$$