The letters of the word MUMMY are arranged in a row randomly. The chance that the letters at the two extremes is M is
1). \(\frac{{3!}}{{5!}}\)
2). \(\frac{{{{\left( {3!} \right)}^2}}}{{5!}}\)
3). \(\frac{{5!}}{{3!3!}}\)
4). 3/20
There are 5 letters in the word with M repeated 3 times.
So, total no. of different ways possible = 5!/3!
If two M are kept fixed at ends, there are 3 remaining letters which can be arranged in 3! Ways.
So, required probability = 3!/(5!/3!)
= (3!)2/5!