Consider the eight letters N, N, J, J, T, T, T, T. How many ways can all eight letters be arranged if no two Ts are adjacent?

[ShanQ]

My first approach
If no Two Ts are adjacent, there will be only 2 arrangements available as below:

T_T_T_T_
or
_T_T_T_T

Take the 1st scenario:
In the empty positions, fit N N J J
Assuming NNJJ are different, then the arrangements are 4! There are 2 N alike, 2 J alike.
Therefore, the number of arrangements are
[math]\frac{4\,!|}{2\,!\times2\,!}=6[/math]
Combine the 1st and 2nd scenarios:
1st scenario OR 2nd scenario → addition rule applies
Therefore, the total number of arrangements: 6+6= 12

My second approach
Step 1
We arrange no-T letters, the number of arrangements are
[math]\frac{4\,!|}{2\,!\times2\,!}=6[/math]
Step 2
Once the non-T letters are arranged, there are 5 possible slots (positions): before the first letter, between the letters, and after the last letter.
For example
Available slots for Ts: _ N _ N _ J _ J _

We choose 4 out of 5 for Ts. The number of arrangements are 5C4 = 5.
Therefore, the total number of arrangements: 6x5=30

I am confused which approach is correct.
Thank you so much for your help.

 

[MaxWong]

The second approach is correct. In the first approach, you forgot to account for cases like TNJTNTJT.

 

[Steven G]

You think that the answer is 12 or 30. Fine. Then try to list them and see which is correct and why one is wrong.