I have a set of values $1,2,3,4,5,6,7,8,9,10$ that I want to place in vectors of length 50 in all permutations.
So far I'm using the permutation formula with $n=50$, $r=10$ as $$ \frac{50!}{(50-10)!} = 3.73*10^{16} $$ Is this correct? It seems inconceivably large... If this is correct, can someone elucidate how I can go about conceptualizing this? I've started with
50 ways to have all 1's and a 2.
50 ways to have all 1's and a 3.
...
50 ways to have all 1's and a 10.
= 9*50 = 450.
then
50 ways to have all 2's and a 1.
50 ways to have all 2's and a 3.
...
50 ways to have all 2's and a 10.
= 9*50 = 450
...
=450*9 = 4,050.
The vector you want to build up can be seen as a number of $50$ digit in base $10$ ($=$ length of the alphabet).
The answer is $10^{50}$.
This is called "sequence with repetition", no permutation.
Similarly, it can be seen as the number of words of $50$ letters given an alphabet of $10$ letters.