In how many ways can the word "kosmos" be arranged so that the same letters do not come together? 
In how many ways can the word "kosmos" be arranged so that the same
   letters do not come together? $(84)$

We have the word $k_1o_1s_1m_1o_2s_2$ - $6$ letters but two letters are of the same kind: $\dfrac{6!}{2!2!}=180$. How to approach the problem further? I am not completely sure I understand why we should divide $6!$ by $2!2!$. Can you clarify this for me? How to approach the problem further?
 A: We wish to find the number of arrangements of the letters of the word KOSMOS in no two identical letters are consecutive.  To do this, we can use the Inclusion-Exclusion Principle.
The word KOSMOS has six letters, so we have six positions to fill.  We can fill two of the positions with O's in $\binom{6}{2}$ ways, two of the remaining four positions with S's in $\binom{4}{2}$ ways, and fill the remaining two positions with the distinct letters K and M in $2!$ ways.  Hence, if there were no restrictions, we would have
$$\binom{6}{2}\binom{4}{2}2! = \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot 2! = \frac{6!}{2!2!}$$
One factor of $2!$ in the denominator represents the number of ways we could permute the two O's within a given arrangement without creating an arrangement that is distinguishable from the given arrangement.  The other factor of $2!$ in the denominator represents the number of ways we could permute the two S's within a given arrangement without creating an arrangement that is distinguishable from the given arrangement.
From these, we must subtract those arrangements in which two O's are consecutive or two S's are consecutive.
Arrangements of the letters of the word KOSMOS in which the two O's are consecutive:  We have five objects to arrange:  K, M, OO, S, S.  Choose two of the five positions for the S's, then arrange the remaining three distinct objects in the remaining three positions. 

 This can be done in $$\binom{5}{2}3!$$ ways.

Arrangements of the letters of the word KOSMOS in which the two S's are consecutive:  We again have five objects to arrange:  K, M, O, O, SS.  By symmetry, there are the same number of arrangements in which the two S's are consecutive as there are arrangements in which the two O's are consecutive.
If we subtract those arrangements in which the two O's are consecutive and those arrangements in which the two S's are consecutive from the total, we will have subtracted those arrangements in which both the two O's are consecutive and the two S's are consecutive twice, once for each way we could have designated one of those pairs of consecutive identical letters as the pair of consecutive identical letters.  We only want to subtract them once, so we must add them back.
Arrangements of the letters of the word KOSMOS in which the two O's are consecutive and the two S's are consecutive:  We have four objects to arrange: K, M, OO, SS.  Arrange the four distinct objects in four positions.

 This can be done in $4!$ ways.

Applying the Inclusion-Exclusion Principle gives the number of admissible arrangements:

 $$\binom{6}{2}\binom{4}{2}2! - 2\binom{5}{2}3! + 4!$$

A: The string is short enough that we can solve the problem by breaking it into three cases


*

*The $o$s and $s$s appear in the order "AABB". In that case, we need to place one of $k$ or $m$ between the first pair, and the other between the second pair to ensure that the numbers aren't consecutive. We are free to swap both the $o$ and $s$, and the $k$ and $m$, so there are $1\times2\times2 = 4$ possibilities for this configuration.

*The $o$s and $s$s appear in the order ABBA. In that case we need to place one of the remaining letters between the two "B"s, and the last letter can go in any of five slots (there are only five slots to avoid double counting). We are free to swap as before, so there are $5\times 2\times2 = 20$ possibilities for this configuration

*The $o$s and $s$s appear in the order "ABAB". In that case we are free to place $k$ and $m$ as we like, since there is no danger of having consecutive letters. We can place the $k$ first in one of 5 slots, and afterwards there are 6 slots for the $m$. We are free to swap the $o$s and $s$s, so there are $5\times 6\times 2 = 60$ possibilities for this configuration
In total, there are thus $4 + 20 + 60 = 84$ possibilities
A: Initially, we have 6 objects with 2 sets of objects with 2 equal members each. The total number of ways = $\dfrac {6!}{2!2!}$
Next, we count those with OO and SS. We first “tie” the objects with the same nature  together to form one unit, and there are two such units. There are 4! Ways to permute these 4 distinct units (namely K, M, O, S). Then, we “unite” the equal objects and let them permute among themselves. There are 2!2! ways. In total, there are 4!2!2! ways.
The no of ways that the same object must not be placed together is then $\dfrac {6!}{2!2!} –4!2!2! = 84$ .
