Number of arrangements no two alike letters together

Question

Find the number of ways in which the letters of the word 'KUTKUT' can be arranged so that no two alike letters are together.

Answer 1

Total letters presents $KK,UU,TT$

$\displaystyle \star\; \large\boxed{K}\boxed{U}\boxed{T}\boxed{-}\boxed{T}\boxed{-}$ we can arrange two remainining letters in $2$ ways in blank space

$\displaystyle \star\; \large\boxed{K}\boxed{U}\boxed{T}\boxed{-}\boxed{-}\boxed{T}$ we can arrange two remainining letters in $2$ ways in blank space

$\displaystyle \star\; \large\boxed{K}\boxed{U}\boxed{-}\boxed{T}\boxed{-}\boxed{T}$ we can arrange two remainining letters in $1$ ways in blank space

Now above state we can arrange $2$ remainining letters in two places is $=2+2+1=5$ ways

Now we can arrange first two letters in each cases $ = 3\times 2 = 6$ ways

So total numbers of arrangements of words is $= 6\times 5 = 30$ ways.

Answer 2

There are $3$ pairs of letters, the simplest way would appear to be through inclusion-exclusion:

(Total ways) - (at least $1$ pair together) + (at least $2$ pairs together) - (all $3$ pairs together), so

$\dfrac{6!}{2^3} - \dbinom31\dfrac{5!}{2^2} + \dbinom32\dfrac{4!}{2} - 3! = 30$

---

Note: I see that my solution is just a compact form of careerbless tool solution 1.

Answer 3

Answer is $30$

Use this careerbless tool (input word=KUTKUT and see generated question $15$)

Answer 4

One way to solve this is to start with arrangements where the first three letters are distinct from each other. There are then $3!=6$ ways to arrange the first three letters, and the only constraint on the final three is that the third and fourth letters are different. This yields $4$ options for the final three letters in your arrangement, or $4\cdot6=24$ total.

Suppose instead that a letter $X$ is repeated in the first three letters of your arrangement. Then it has to occur in the first and third spots in your arrangement. Once you choose that letter, and also the letter $Y$ going in the second spot (which has no restrictions other than it be distinct from $X$), the remaining three letters in the arrangement are determined. The third letter $Z$ is duplicated and appears in the fourth and sixth spots, and the letter $Y$ appears in the fifth spot. Hence, we have $3$ choices for $X$ and $2$ choices for $Y$, for $3\cdot2=6$ total choices in this case.

Adding the cases together, we have $6+24=\boxed{30}$ total arrangements where no two alike letters are together.