Lets say a car plate has 6 positions, three of them are numbers and 3 of them are alphabetic.
e.g. ABC 123. I need to figure out how many possible permutations can occur using all the numbers and alphabets BUT i cannot use any number or alfabet twice in an permutation eg. AAB 123 or ABC 112 is forbidden.
The Attempt at a Solution
first i split the problem, first the numbers then the alfabet:
so N1 is the number of numbers i can use which is 10 obviously. 0-9
K is the number of positions which is 3.
so N2 is the number of alfabets which is 26, K is the number of positions 3.
(N1!/(N1-K))*(N2!/(N2-K)!) = (10!/(10-3)!)*(26!/(26-3)!)
is the answer correct?