If the letters of the word ASSASSINATION are arranged at random. Find the Probability that (a) All A’s are not coming together (b) No two A’s are coming together.

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If the letters of the word ASSASSINATION are arranged at random. Find the Probability that
(a) All A’s are not coming together
(b) No two A’s are coming together.

CBSE | Class 11 | Maths | NCERT Exemplar | Probability

If the letters of the word ASSASSINATION are arranged at random. Find the Probability that
(a) All A’s are not coming together
(b) No two A’s are coming together.

Solution:

The given word is ASSASSINATION

Total no. of letters in ASSASSINATION $=13$

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

$\therefore$ The total no. of ways in which these letters can be arranged

$=\mathrm{n}(\mathrm{S})=\frac{13 !}{3 ! 4 ! 2 ! 2 !}$
(a)All A’s are not coming together
We firstly need to find the probability that all A’s are coming together
If all the A’s are coming together, so
Now the no. of letters is $1+10=11$
$\therefore$ No. of words when all A’s come together $=\frac{11!}{4!2!2!}$
$\therefore$ Probability when all $\mathrm{A}^{\prime}$ s come together $=\frac{\frac{11 !}{4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
$=\frac{11 !}{4 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !}$
We can also write the above equation as
$=\frac{11 ! \times 3 !}{13 \times 12 \times 11 !}$
$=\frac{3 \times 2 \times 1}{13 \times 12}$
Now simplify
$=\frac{1}{26}$
Now, Probability (all A’s does not come together) = 1 – Probability (all A’s come together)
$=1-\frac{1}{26}$
$=\frac{25}{26}$

(b) No two A’s are coming together
We firstly need to arrange the alphabets except $A^{\prime} s$ ,
$\therefore$ The no. ofways of arranging all alphabets except $\mathrm{A}^{\prime} \mathrm{s}=\frac{10 !}{4 ! 2 ! 2 !}$
Between these alphabets there are 11 vacant places.
Total no. of A’s in the word ASSASSINATION are 3 .
$\therefore 3 \mathrm{~A}^{\prime} \mathrm{s}$ can be placed in 11 place in ${ }^{11} \mathrm{C}_{3}$ ways
$=\frac{11 !}{3 !(11-3) !}$
$\begin{array}{l} =\frac{11 !}{3 !(11-3) !} \\ =\frac{11 !}{3 ! 8 !} \end{array}$
$\therefore$ The total no. of words when no two A’s together
$=\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !}$
$\therefore \text { Required Probability }=\frac{\frac{11 !}{3 ! 8 !} \times \frac{10 !}{4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
We can also write the above equation as
$=\frac{11!}{3!8!}\times \frac{10!}{4!2!2!} \times \frac{3!4!2!2!}{13!}$
$=\frac{11!\times10\times9\times8!}{8!\times13\times12\times11!}$
$=\frac{10\times9}{13\times12}$
$=\frac{15}{26}$