Question

If the letters of the word ASSASSINATION are arranged at random. Find the probability that
(i) Four $S's$ come consecutively in the word
(ii) Two $I's$ and two $N's$ come together
(iii) All $A's$ are not coming together
(iv) No two $A's$ 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 $3A's$, $4S's$, $2I's$, $2N's$, $1T's$ and $1\, O's$.
(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 $= \frac{10! \times 3! \times 4! \times 2! \times 2!}{3!2!2! \times 13!}$
$= \frac{10! \times 4!}{13!}=\frac{4!}{13 \times 12 \times11}$
$= \frac{24}{1716} = \frac{2}{143}$
(ii) If $2I’s$ and $2N’s$ come together, then there are $10$ alphabets.
$\therefore $ Number of words when $2I’s$ and $2N's$ 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 $A's$ come together $= \frac{11!}{4!2!2!}$
$\therefore $ Probability when all $A's$ 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 $A's$ does not come together $= 1-\frac{1}{26}=\frac{25}{26}$
(iv) If no two $A's$ are coming together, then first we arrange the letters except $A's$

All the letters except $A's$ are arranged in $= \frac{10!}{4!2!2!}$ ways
There are $11$ vacant places between these letters.
Therefore, $3A's$ 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 $= \frac{11! \times 10!}{3!8!4!2!2!}\times \frac{4!3!2!2!}{13!}$
$= \frac{10!}{8! \times 13\times 12}= \frac{10\times 9}{13\times 12}$
$= \frac{15}{26}$