How to calculate the number of permutations and combinations if k is equal to n? Say the question is 

How many unique ways are there to arrange the letters in the word FANCY?

The formula I use for permutations is n! / (n - k)! 
Combinations formula I use is n! / ( k! * (n - k)! )
In this case, I would be dividing by zero in both cases. What universal formula can I use for situations where k might be equal to or larger than n?
Thank you
 A: By definition, $0!=1$. There are a number of different justifications for this. One of them is that this choice makes the formulas you quote give the right answer. One can also argue that there really is exactly one way to choose $0$ objects from $n$: just say no in turn to each of them.
When $k$ and $n$ are non-negative integers, and $k\gt n$, we have two choices for $\binom{n}{k}$. We could say it is undefined. But it is more convenient to adopt the convention that in that situation, $\binom{n}{k}=0$. Again one could give an informal justification, there are $0$ ways to choose $5$ people from $3$. But the real reason for the convention is that it makes some formulas less messy. 
A: For the word FANCY:

*

*The number of members we have is n=5 (F,A,N,C,Y)


*The number of selection from the member is k=5 (as well)
So n!/(n-k)! = 5!/(5-5)! = 5!/0! = 5! (as 0! = 1)
The final result is 120
Another way of doing this is consider them as slots:

*

*for slot 1 we have 5 members to choose from {F,A,N,C,Y}

*for slot 2 we have 4 members left to choose from ({F,A,N,C,Y} - {whatever chosen at slot1})

*for slot 3 we have 3 members left to choose from ({F,A,N,C,Y} - {whatever chosen at slot1 and slot2})

*for slot 2 we have 2 members left to choose from ({F,A,N,C,Y} - {whatever chosen at slot1, slot2 and slot3})

*for slot 3 we have 1 members left to choose from ({F,A,N,C,Y} - {whatever chosen at slot1 and slot2})

Multiply all the possible options we have from each slot we got
    1 x 2 x 3 x 4 x 5

which is 5!
If you are interested in understand combination and permutation more, this might be a very interesting link: https://www.mathsisfun.com/combinatorics/combinations-permutations.html
