Question

If the letters of the word ASSASSINATION are arranged at random. Find the probability that

Answer 1

We have 13 letters in which $3A{}^{\prime }4S{}^{\prime },2i{}^{\prime }s,2N{}^{\prime }s1T10$.

Total number of ways these letters can be arranged=$n\left(s\right)=\frac{13!}{3!4!2!2!}$

if fourS's come consecutively in the word Consider 4 Ss as as a single until,then arrangement is like this(SSSS),A,A,A,I,I,N,N,T,O

Thus,we have total 10 units out of which (SSSS),3A's,2I's,2N's,1T&10


$\therefore$ Total arrangements where S's together,

$n\left(E\right)=\frac{10!}{3!\times 2!\times 2!}$

$\therefore$ Required probability


$=\frac{n\left(E\right)}{n\left(S\right)}=\frac{\frac{10!}{3!\times 2!\times 2!}}{\frac{13!}{3!4!2!2!}}$

$\frac{10!\times 4\times 3\times 2}{13\times 12\times 11\times 10!}$

$=\frac{2}{143}$