What is the probability that a random arrangement of the letters in the word `SEVEN' will have both E's next to each other?

Yep I agree with Ninja and CPhill  :)

24/60 = 2/5

Too lazy to show work. Ask if you need it.

5!/2!3! = .083 or 1/12.

There 120 = 5! ways of aranging the 5 letters in "Seven". You want to know the number of ways out of 120 that you can occupy 2 spaces out of 5, leaving 3 left over.

Source: "Probability, An Introduction" by Samuel Goldberg.

Here's my take:

Total identifiable "words"  generated by the arrangements of the letters in"SEVEN"   =

5!  / 2!  =   120  / 2   =   60

Possible identifiable "words"  formed by having "E's"  next to each other  = the "E's"  can occupy any of 4 positions....and for each of these......the oher letters can be arranged in 3!  = 6 ways  =  4 * 6    = 24

So.....the probability =   24/ 60    =    2 / 5

I agree with NinjaAnswer  !!!!

Well....three "great minds"  couldn't possibly be wrong.......LOL!!!!!

And NinjaAnswer is ONLY 12 years old!!.

Lol thanks!