<s>[INST] The number of positive integral solutions of $abc =$ $30$ is 
The number of positive integral solutions of $abc =$ $30$ is

My attempt
Factors of $30$ are $3 \cdot 5 \cdot 2$.
Therefore $a$ will have three choices, similarly $b$ & $c$ will have $2$ choices & $1$ choice.
So,it's answer must be $3 \cdot 2 \cdot 1$
But answer in my textbook is given is $3 \cdot 3 \cdot 3$.
.It states that"$2$ can be assigned to either $a$ or $b$ or $c$".  Similarly, each of $3$ and $5$ can be assigned in $3$ ways.  Thus, the number of solutions is $3 \cdot 3 \cdot 3$ 
According to me, if $a$ is already assigned to $2$ than how can other letters ($b$ or $c$) have same value as that of $a$.
Please explain me where am I wrong?
 [/INST] The texbook is right.
Start with $1\cdot1\cdot1$ and assign $2$, say to $b$ and get $1\cdot2\cdot1$.
Now assign $3$, say to $a$ and get $3\cdot2\cdot1$.
Then assign $5$ to $a$ again, giving $15\cdot2\cdot1$.
On every stage, you have three choices.
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