Number of permutations with repetition I have to form a permutation of N elements with each element less than or equal to  M.
I need to count the number of unique permutations possible with 2 of the elements fixed as 2 and 3. 
Note that each number can be used multiple times and 2 and 3 can occur anywhere in the permutation.
I came up with the formula : 
N*(N-1)* pow(M , N-2 ).
N places for 3 to occur , hence there are N-1 places for 2 to occur followed by M choices for the remaining N-2 places.
However it is resulting in over-counting.
for N=3 M=4 :
fix first place for 3 and second place for 2:
3 2 1
3 2 2
3 2 3
3 2 4
fix first place for 3 and third place for 2:
3 1 2
3 2 2
3 3 2
3 4 2
Already there is over-counting of the permutation 3 2 2 in my method.
How do I determine the total unique permutations? 

I'm reading that the combination must include at least one 2 and at least one 3 somewhere in it. 
Suppose you just consider one of required inclusions:
Constraint A: the combination must include at least one 2
How many of the total possibilities without Constraint A are removed by this constraint?
Total combinations $C_T = M^N$
with constraint $A$ only, $C_A = M^N - (M-1)^N$
for the other constraint:
Constraint B: the combination must include at least one 3
with constraint $B$ only, $C_B = M^N - (M-1)^N$
However with both constraint we need inclusion-exclusion to make sure we don't double-count excluded possibilities:
Both constraints: 
$C_{AB} = M^N - 2(M-1)^N + (M-2)^N$
