permutation of letters in "Combinatorics" w/ no 2 consec. letters identical

Find the number of permutations of letters of the word ’COMBINATORICS’ where 2 consecutiveletters are never identical.

indianatoms said:

Find the number of permutations of letters of the word ’COMBINATORICS’ where 2 consecutive letters are never identical.

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This is a problem of separators. We have three letters that repeat and must be separated from one another. So consider the seven letters that do not repeat.
\(\displaystyle \mathop {\_\_}\limits_1 {\text{M}}\mathop {\_\_}\limits_2 {\text{B}}\mathop {\_\_}\limits_3 {\text{N}}\mathop {\_\_}\limits_4 {\text{A}}\mathop {\_\_}\limits_5 {\text{T}}\mathop {\_\_}\limits_6 {\text{R}}\mathop {\_\_}\limits_7 {\text{S}}\mathop {\_\_}\limits_8\)
As you can see they create eight places to put the two C's such as follows:
\(\displaystyle \mathop {\_\_}\limits_1 {\text{M}}\mathop {\_C\_}\limits_2 {\text{B}}\mathop {\_\_}\limits_3 {\text{N}}\mathop {\_\_}\limits_4 {\text{A}}\mathop {\_\_}\limits_5 {\text{T}}\mathop {\_\_}\limits_6 {\text{R}}\mathop {\_\_}\limits_7 {\text{S}}\mathop {\_C\_}\limits_8\)
That can be done in \(\displaystyle \dbinom{8}{2}=28\) ways. But looking that, we now have ten places to put the two i's. Can be done in \(\displaystyle \dbinom{10}{2}=~?\) ways. Do it one more time to place the O's.
How many ways can we arrange the original seven separators?