# Count of all possible combinations of length $K$ for a given string of $N$ letters

I need to find the count of all possible combinations of length $$K$$ for a given string $$N$$. The characters cannot repeat.

For example, if the given string is $$a,b,c,d,\quad N=4$$

The possible number of combinations of length $$K=2$$ will be $$12$$: $$ab,ac,ad,bc,bd,cd,ba,ca,cb,da,dc,db$$

The possible number of combinations of length $$K=3$$ will be $$24$$: $$abc,abd,acb,acd,adb,adc,bac,bad,bcd,bca,bda,bdc,cab,cad,cba,cbd,cda,cdb,dab,dac,dba,dbc,dca,dcb$$

Is the formula $$\dfrac{N!}{(N-K)!}$$ correct?

Your formula is correct, and here is why:

There are $$\displaystyle{N \choose K} = \frac{N!}{(N-K)!K!}$$ ways to pick a set of $$K$$ non-repeating characters out of $$N$$ characters.

Also, once you have those $$K$$ different characters picked, you can order them in $$K!$$ ways.

So, there are

$$\frac{N!}{(N-K)!K!} \cdot K! = \frac{N!}{(N-K)!}$$

possible strings of $$K$$ different characters chosen from a group of $$N$$ characters.

Your result is correct. The number is $$\frac{N!}{(N-K)!}.$$

Indeed there are $$N$$ ways to choose the first letter, $$N-1$$ to choose the second one and so on until $$N-K+1$$ ways to choose the $$K$$-th letter. Multiplication of these numbers gives the result.