For example assume we have the following word "dummybriaan", and we want to arrange without having two m:s after eachother.
then we rewirte the word with out the m:s and we get "duybriaan" which can be arranged in $\left(\frac{9!}{2!}\right)$.
Now we find the way to place the m:s in the word, _d_u_y_b_r_i_a_a_n_, which gives us 10 postions => $10\choose2$
and the totalt arrangements is $\left(\frac{9!}{2!}\right)$ * $10\choose2$.
but what if we want to arrange without repetition, meaning as we did for m:s we have now to include a:s as well ?
There are eleven letters in DUMMYBRIAN. We can arrange the seven distinct letters B, D, I, N, R, U, Y in a row in $7!$ ways. That leaves us with eight spaces in which to place the two A's, six between successive letters and two at the ends of the row. We have two options, place both letters in the same space or place them in two separate spaces.
Case 1: Place both A's in the same space.
There are eight ways to choose the space. We now have nine letters: A, A, B, D, I, N, R, U, Y. There are now ten spaces in which we can place an M, the eight spaces between successive letters and the two ends of the row. Since we wish to ensure that no two consecutive letters are the same, we must place an M between the two A's and place the other M in one of the other nine spaces. The number of such arrangements is $$7!\binom{8}{1}\binom{9}{1}$$
Case 2: Place the two A's in different spaces.
There are $\binom{8}{2}$ ways to place the two A's in two of the eight spaces created by arranging the letters B, D, I, N, R, U, Y in a row. We now have an arrangement of nine letters, which creates ten spaces in which to place the two M's. Since the M's must be separated, we must choose two of these ten spaces for the M's, which can be done in $\binom{10}{2}$ ways. The number of such arrangements is $$7!\binom{8}{2}\binom{10}{2}$$
Total: Notice that the sequence of letters AMA is only possible in the first case since there is always at least one letter other than M between the two A's in the second case. Hence, the two cases are disjoint. Therefore, the number of arrangements in which neither the two A's nor the two M's are consecutive is $$7!\binom{8}{1}\binom{9}{1} + 7!\binom{8}{2}\binom{10}{2}$$