Consider generating a 4-length code using only 3 digits. The possible number of permutations is 34=81. However what if all 3 digits had to be used. How many permutations exist? In other words only one duplication of a digit is allowed, but it could be any of the digits.
Drawing a tree I can see some sort of pattern for n=3 r=4 and n=4 r=5 but I can't get it to gel and extrapolate to n=3 r=5 where two duplications or a triple is allowed and so on.
Is there a generalised formula that allows this to be calculated?
TIA
Update - to put it another way if n is the number of elements and r is the number to be selected then if I write the permutation as P(n r) then how to calculate P(3 4)? Further I know that P(4 4) =24 and that P(3 4) is 36, but P(2 4) is 14 and of course P(1 4) is 1. Lowering n then P(3 3) is 6, but P(2 3) is 8. So beyond how to calculate P(n-1 n) can it be shown that P(n-1 n) > P(n-2 n) for all cases of n?
