Derangement with repetition

There are $\binom{4}{2}=6$ ways of placing the two Os in new positions. For each such position, there are two of the other four letters whose original position has been chosen, and so can go in any of the remaining places, while the other two letters could, but must not, go in their original positions.

Let us count the number of ways of rearranging XYZW such that neither X nor Y is in its original position. There are $3!$ ways of rearranging the numbers so that X is in its original position, $3!=6$ ways of rearranging the numbers so that Y is in its original position, and $2!=2$ ways of rearranging the numbers so that both X and Y are in their original position. Thus there are $6+6-2=10$ ways of rearranging the numbers so that either X or Y is in its original position, and hence $4! - 10 = 14$ ways of rearranging the numbers so that neither X nor Y is in its original position.

Thus there are $6 \times 14 = \boxed{84}$ ways of rearranging the letters in FOODIE so that no letter is in its original position.

There are $4\cdot 3 = 12$ ways to fill two O's places with two of the 4 remaining letters. After that, we are left with two O's and two other letters . Now, one of those two letters can be placed either on one of the free places (which came to be free because we've used letters from these positions for filling O's positions) or on the place of the other letter of the two. In the first scenario, after placing one letter on one of two free places, there are two possible places for the other letter - second free place or the place of the first letter - making total of $4$ combinations. In the second scenario, the other letter can be placed on all of remaining three positions - making total of $3$ combinations. In the end, we fill remaining two places with Os. Hence, there are $12\times (4+3) = 84$ ways to perform desired rearranging.

Python:

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F = ["F","O","O","D","I","E"]
L = list(itertools.permutations(F))  #all permutations of this list.
count = 0
for word in L:
    if (derange(F,word)):  #defined appropriately
        count +=1   #count is double-counting, since the iterator treats the two 'O's as distinct characters  
return(count // 2)

>84

The number of Derangement with n distinct objects and p identical objects $D_{n, p}$ is:

$\displaystyle D_{n,p}={n \choose p}\sum_{k=0}^{n-p} {n-p \choose k}(-1)^k(n-k)!$

For $n=4$ and $p=2$ we have

$D_{4, 2}=84$