If all the letters of the word 'SENSELESSNESS' are arranged in all possible ways and an arrangement among…
If all the letters of the word 'SENSELESSNESS' are arranged in all possible ways and an arrangement among them is chosen at random then, the probability that all the E's come together in that arrangement is
$\frac{1}{990}$
$\frac{2}{143}$
$\frac{1}{120}$
$\frac{1}{429}$
Solution
Total words can be made out of 'SENSELESSNESS'
$=\frac{13!}{6!4!2!}=180180$ Total words can be made with all the $E$ 's together
$=\frac{(13-3)!}{6!2!}=2520$
$\therefore$ Required probability $=\frac{2520}{180180}=\frac{2}{143}$.