# If the letters of the word ASSASSINATION

Question:

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

(a) Four S’s come consecutively in the word

(b) Two I’s and two N’s come together

(c) All A’s are not coming together

(d) No two A’s are coming together.

Solution:

Given word is ASSASSINATION

Total number of letters in ASSASSINATION is 13

In word ASSASSINATION, there are 3A’s, 4S’s, 2I’s, 2N’s, 1T’s and 1O’s

Total number of ways these letters can be arranged =

$\mathrm{n}(\mathrm{S})=\frac{13 !}{3 ! 4 ! 2 ! 2 !}$

(a) Four S's come consecutively in the word

If $4 S^{\prime}$ s come consecutively then word ASSASSINATION become.

$\therefore \mathrm{n}(\mathrm{E})=\frac{10 !}{3 ! 2 ! 2 !}$

$\therefore$ Required Probability $=\frac{\frac{10 !}{3 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$

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

The above equation can be written as

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

$=\frac{4 \times 3 \times 2 \times 1}{13 \times 12 \times 11}$

On simplifying we get

$=\frac{2}{143}$

(b) Two I's and two N's come together So, now numbers of letters is $1+9=10$

$\therefore \mathrm{n}(\mathrm{E})=\frac{4 !}{2 ! 2 !} \times \frac{10 !}{3 ! 4 !}$

$\therefore$ Required Probability $=\frac{\frac{4 ! 10 !}{3 ! 4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$

The above equation can be written as

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

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

$=\frac{4 \times 3 \times 2 \times 1}{13 \times 12 \times 11}$

On sımplityıng we get

$=\frac{2}{143}$

On simplifying we get

$=\frac{2}{143}$

(c) All A's are not coming together

Firstly, we find the probability that all A's are coming together If all A's are coming together then

$\therefore$ Number of words when all A'scome together $=\frac{11 !}{4 ! 2 ! 2 !}$

$\therefore$ Probability when all A's come together $=\frac{\frac{11 !}{4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$

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

The above equation can be written as

$=\frac{11 ! \times 3 !}{13 \times 12 \times 11 !}$

$=\frac{3 \times 2 \times 1}{13 \times 12}$

On simplifying we get

$=\frac{1}{26}$

Now, $\mathrm{P}$ (all $\mathrm{A}^{\prime}$ s does not come together) $=1-\mathrm{P}$ (all $\mathrm{A}^{\prime}$ s come together)

$=1-\frac{1}{26}$

$=\frac{25}{26}$

(d) No two A's are coming together First we arrange the alphabets except A's

$\therefore$ Number ofways of arranging all alphabets except $\mathrm{A}^{\prime} \mathrm{s}=\frac{10 !}{4 ! 2 ! 2 !}$

There are 11 vacant places between these alphabets. Total A's in the word ASSASSINATION are 3

$\therefore 3$ A's can be placed in 11 place in ${ }^{11} C_{3}$ ways

$=\frac{11 !}{3 !(11-3) !}$

$=\frac{11 !}{3 ! 8 !}$

$\therefore$ Total number of words when no two A's together

$=\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !}$

$\therefore$ Required Probability $=\frac{\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$

The above equation can be written as

$=\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !}$

$=\frac{11 ! \times 10 \times 9 \times 8 !}{8 ! \times 13 \times 12 \times 11 !}$

$=\frac{10 \times 9}{13 \times 12}$

On simplifying we get

$=\frac{15}{26}$