Question

If the letters of the word ASSASSINATION are arranged at random. Find the probability that
(i) Four $S^{\prime }s$ come consecutively in the word
(ii) Two $I^{\prime }s$ and two $N^{\prime }s$ come together
(iii) All $A^{\prime }s$ are not coming together
(iv) No two $A^{\prime }s$ are coming together

	(i)	(ii)	(iii)	(iv)
(a)	$\frac{2}{143}$	$\frac{2}{143}$	$\frac{25}{26}$	$\frac{15}{26}$
(b)	$\frac{25}{26}$	$\frac{15}{26}$	$\frac{2}{143}$	$\frac{2}{143}$
(c)	$\frac{15}{26}$	$\frac{25}{26}$	$\frac{2}{143}$	$\frac{2}{143}$
(d)	$\frac{2}{143}$	$\frac{25}{26}$	$\frac{2}{143}$	$\frac{15}{26}$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (A)

Total number of letters in the word ASSASSINATION are $13$.
Out of which $3A^{\prime }s$, $4S^{\prime }s$, $2I^{\prime }s$, $2N^{\prime }s$, $1T^{\prime }s$ and $1O^{\prime }s$.
(i) If four $S’s$ com e consecutively, then we consider these $4S’s$ as $1$ group.

$\therefore$ Number of words when all $S’s$ are together $=\frac{10!}{3!2!2!}$
Also, total number of words using letter of the word
ASSASSINATION $=\frac{13!}{3!4!2!2!}$
$\therefore$ Required probability $=\frac{10!\times 3!\times 4!\times 2!\times 2!}{3!2!2!\times 13!}$
$=\frac{10!\times 4!}{13!}=\frac{4!}{13\times 12\times 11}$
$=\frac{24}{1716}=\frac{2}{143}$
(ii) If $2I’s$ and $2N’s$ come together, then there are $10$ alphabets.
$\therefore$ Number of words when $2I’s$ and $2N^{\prime }s$ come together
$=\frac{10!}{3!4!}\times \frac{4!}{2!2!}$
$\therefore$ Required probability $=\frac{\frac{10!4!}{3!4!2!2!}}{\frac{13!}{3!4!2!2!}}$
$=\frac{4!10!}{13!}=\frac{2}{143}$
(iii) Number of words when all $A^{\prime }s$ come together $=\frac{11!}{4!2!2!}$
$\therefore$ Probability when all $A^{\prime }s$ come together
$=\frac{\frac{11!}{4!2!2!}}{\frac{13!}{4!3!2!2!}}=\frac{11!\times 3!}{13!}$
$=\frac{6}{13\times 12}=\frac{1}{26}$
$\therefore$ Required probability when all $A^{\prime }s$ does not come together $=1-\frac{1}{26}=\frac{25}{26}$
(iv) If no two $A^{\prime }s$ are coming together, then first we arrange the letters except $A^{\prime }s$

All the letters except $A^{\prime }s$ are arranged in $=\frac{10!}{4!2!2!}$ ways
There are $11$ vacant places between these letters.
Therefore, $3A^{\prime }s$ can be placed in $11$ places in ${}^{11}{C}_3=\frac{11!}{3!8!}$ ways
$\therefore$ Total number of words when no two A's are together
$=\frac{11!}{3!8!}\times \frac{10!}{4!2!2!}$
$\therefore$ Required probability $=\frac{11!\times 10!}{3!8!4!2!2!}\times \frac{4!3!2!2!}{13!}$
$=\frac{10!}{8!\times 13\times 12}=\frac{10\times 9}{13\times 12}$
$=\frac{15}{26}$