basic permutations problem doubt If I want to find out how many possible combinations are there of $3$ traffic lights getting turned on, then the answer is $27$.  We can get the answer by $3^3 = 27$.  If we use a factorial of $3$, like in permutations, we get $6$, which is incorrect.
So why can't we use the same logic to find the answer of the below question: "How many ways are there to arrange the letters in the word ASCENT?"
As there are $6$ different alphabets filling $6$ different positions, so shouldn't the answer be $6^6 = 46656$?.
Why is the permutations formula, $6$ factorial correct in this scenario?
$6! = 720$.
I am unable to understand this. If anyone could please help me, would be a great help.
Thanks
 A: It is very simple because the restriction in both the cases is completely different.
In the first case all the three lights can take any color, which means repetitions are allowed, so there are no restrictions. Therefore, we can say that for every light we have $3$ choices, straight forward applying the fundamental principle of counting we have number of choices:
$3 \cdot 3 \cdot 3 = 3^3$.
But in the case of permutations of the letters of the word ASCENT, there is a restriction which one must identify very logically, i.e. repetitions can not take place. Just simply think that you cannot use a letter again (e.g. you cannot make word like AAAAAA).  Therefore, first we have $6$ choices (A,C,E,N,S,T). After that since we have filled one place, one letter has been used, so now we have $5$ choices. Then we will have $4$, then $3$, then $2$, then $1$. Applying fundamental principle of counting:
$6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 6!$.
So there you go. Long said short when repetitions are allowed the permutations are : $m^n$, where $m$ is number of choices and $n$ is the number of available places.  When repetitions are not allowed the permutations are $n!$.
Hope this helps. Keep lovin' MATHS!
