Find the no. Of arrangements of letters of the world SALOON if two 'O's do not come together?

If the 6 letters in SALOON were all different, there'd be #6! = 720# ways to arrange the letters.

However, there is a duplicate O, and so the total number of unique ways to arrange the letters is:

#(6!)/(2!)=720/2=360#

(we divide out the duplicates by the number of ways the Os can be internally arranged, which in this case is #2!#)

Of those ways the letters can be arranged, how many involve not having the 2 Os together? We can figure this out by either looking at the different positions of the Os that satisfy this condition, or we can subtract out the number of ways that don't (from the total number of ways). I'll do it that second way.

We can put the 2 Os in positions 1, 2; 2, 3; ... 5, 6, or 5 positions in total.

We've already divided out the duplicates from the two Os in the 360, so we won't multiply the 5 position number by 2 (the way we would if the letters were unique).

Once the Os are positioned, there are 4 places for the remaining 4 letters to go, so that's #4! = 24# ways.

All told, that's

#5xx24 = 120# ways that are disallowed. This gives:

#360-120 = 240# ways that are allowed.

~~~~~

To verify, we can work the question the other way. We can place the Os in positions

1, 3; 1, 4; 1, 5; 1,6
2, 4; 2, 5; 2, 6
3, 5; 3, 6
4, 6

or 10 ways. There are #4! =24# ways to place the remaining letters, or

#10xx24=240# ways in total that are allowed