A machine has $9$ switches. Each switch has $3$ positions. How many different settings are possible? A machine has $9$ switches. Each switch has $3$ positions. 
$(1)$ How many different settings are possible?
Each switch has $3$ different settings and we have $9$ total. So,
$3^9=19,683$
Now, the problem I am facing is what the heck the second part is asking.
$(2)$ Answer $(1)$ if each position is used $3$ times. How would you interpret this? I see it as; if we have a certain position for the $9$ switches, say, ABCABCABC then that position is accounted for $2$ additional times $(3$ times total$)$. Thus, we would multiply part $(1)$ by $3$ because we are simply tripling our possibilities?
Any ideas on how part $(2)$ should be approached would be great!
 A: Hint for 2: How many different words can you spell with the letters AAABBBCCC?
A: The answer to (2) is supposed to be less than the answer to (1), because instead being free to pick any of the 3 positions for each of the 9 switches, you can only pick combinations that have three switches in the A position, three in the B position, and three in the C position. So ABCABCABC works, AAABBBCCC works, but AAAAABBCC is not an option.
A: Select one of the three positions. How many ways can you choose 3 of the 9 switches with that position? After that, how many ways can you choose 3 of the 6 switches in one of the two different remaining switches? Finally, how many ways can you choose the remaining 3 switches for the last position? 
What are the total number of possibilities? 
A: Actually it is very simple. Position as was noted by MPW is just value 1,2,3 for a switch. So in (2) we have all combinations of 3 ones,twos, and thirds. The answer is just $9!/6^3$ because  the first 1 can be set on 9 position, the second on 8, etc. After this since we count any 3 same digits 6 times we have to divide it by 6^3.  This is how I see the picture,
So IMHO answer is $9!/6^3=1680$.
A: You're off to a good start.
If each position is used exactly three times you do indeed wish to count the permutations of a string such as “AAABBBCCC”.  This is the multiset permutation: $$\dbinom{9}{3, 3, 3}=\dfrac{9!}{3!3!3!}$$
There are $9!$ ways to arrange $9$ symbols.  However, many of these will be equivalent, because the order of identical symbols is not important.  There are $3!$ equivalent ways to arrange three identical symbols.  Thus there are $\frac{9!}{3!3!3!}$ distinct ways to arrange $3$ groups of $3$ kinds of symbols.
