Find the cardinality of set $\{(x,y,z): x,y,z\text{ are positive integers and }xyz=108\}$
my attempt:
I write all factors of $108$.
$$1,2,3,4,6,9,12,18,27,36,54,108.$$ First observe that $xyz $ has only two degree of freedom. So its enough to choose $x,y$ . If I choose $x=1$ there are $12$ ways of choosing $y$. If choose $x$ as $2$ there are $11$ ways ...and so on.
So total number of ways $=1+2+...12=78$. But its give in my book that $60$ is answer. Where am I wrong?
On
$$(2^{j_1} 3^{k_1})(2^{j_2}3^{k_2})(2^{j_3}3^{k_3}) = 108 = 2^2 \cdot 3^3$$
where $\{ j_i, k_i\} \ge 0$ and by the Prime Factorization Theorem we have $j_1 + j_2 + j_3 = 2$ and $k_1 + k_2 + k_3 = 3$.
How many ways can you solve those equations, up to equivalent permutations of the factors? $12$
Start with the first (smallest) factor $(2^0 3^0)$ and increment others for the first "row"; then $(2^1 3^0)$ and increment others for the second "row"; then $(2^0 3^1)$ and so on... for the third "row." There is no fourth "row"...
Here are the sets, each in ascending order, themselves ordered:
$108=2^2 \cdot 3^3$, so $x,y,z$ must be of the form $2^i3^j$ such that the sum of the $i$'s (where $0\le i\le 2$) is $2$ and the sum of the $j$'s (also integers starting from $0$) is $3$. How many solutions do $i_x+i_y+i_z = 2$ and $j_x+j_y+j_z = 3$ have in integers $\ge 0$? All combinations of these solutions are the solutions you want.
There are $\binom{n+k-1}{k-1}$ non-negative integer solutions to $x_1+ \ldots +x_k =n$ (see wikipedia, e.g.) and so for $3$ variables and $n=2$ we get $\binom{2+3-1}{2} = \binom{4}{2}=6$ and for $n=3$ we get $\binom{5}{2} = 10$. So combined we get $6 \times 10=60$ solutions for these independent equations.