Permutations help please.

The total arrangements is 8!, but the total arrangements where there are 2 same letters next to one another is 7! * 3, since it is 7! for each letter, A, L, and S. Then, there are 6! * 3 scenarios where two of the same letters are next to one another, either A and L, A and S, or L and S. Then, there are 5! arrangements where all letters are next to the other same letter. We then subtract everything, giving us 8! - (7!*3) - (6!*3) - 5!, or 22920.

Unfortunately, that's wrong, but nice try!

There are in total

$\frac{8!}{2!2!}$ total arrangements of letters

considering arrangements of letter where the two 'L's are right next to each other, we have

$\frac{7!}{2!}$ total arrangements

and we have the same amounts of arrangements for the two 'S's being rigth next to each other

So therefore, we have

$\frac{8!}{2!2!} - 2 * \frac{7!}{2!} + 6!$ because we overcount a bit

so the answer is $\fbox{4760}$

cycle:c=if((a[i]!=a[j] and a[j]!=a[k] and a[k]!=a[l] and a[l]!=a[m] and a[m]!=a[n] and a[n]!=a[r] and a[r]!=a[s]), goto end, goto next);end:print(a[i],a[j],a[k],a[l],a[m],a[n],a[r],a[s]),", ",;p=p+1;next:i=i+8;j=i+1;k=j+1;l=k+1;m=l+1;n=m+1;r=n+1;s=r+1;if(s< count a, goto cycle, 0);
;print">Total =", p

This computer code gives a total of: 2,220 permutations, where no 2 of the same letters are next to each other.

By the way, there are: 8! / 2!2!2! =5,040 permutations in total.

Edit: Guest who posted

cycle:c=if((a[i]!=a[j] and a[j]!=a[k] and a[k]!=a[l] and a[l]!=a[m] and a[m]!=a[n] and a[n]!=a[r] and a[r]!=a[s]), goto end, goto next);end:print(a[i],a[j],a[k],a[l],a[m],a[n],a[r],a[s]),", ",;p=p+1;next:i=i+8;j=i+1;k=j+1;l=k+1;m=l+1;n=m+1;r=n+1;s=r+1;if(s< count a, goto cycle, 0);
;print">Total =", p

was correct!

TYSM!!