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

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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

CBSE | Class 11 | Maths | NCERT Exemplar | Probability

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

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) Four S’s come consecutively in the word
If $4 \mathrm{~S}^{\prime}$ s comes consecutively then the word ASSASSINATION become.
Therefore, now the no. of letters is $1+9=10$
$\begin{array}{l} \therefore \mathrm{n}(\mathrm{E})=\frac{10 !}{3 ! 2 ! 2 !} \\ \therefore \text { The required Probability }=\frac{\frac{10 !}{3 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}} \\ =\frac{10 !}{3 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !} \end{array}$
We can write the above equation as
$\begin{array}{l} =\frac{10 ! \times 4 !}{13 \times 12 \times 11 \times 10 !} \\ =\frac{4 \times 3 \times 2 \times 1}{13 \times 12 \times 11} \end{array}$
$=\frac{2}{143}$
(b) Two I’s and two N’s come together
Therefore, now no. of letters is $1+9=10$
$\therefore n(E)=\frac{4 !}{2 ! 2 !} \times \frac{10 !}{3 ! 4 !}$
$\therefore$ The required Probability $=\frac{\frac{4 ! 10 !}{3 ! 4 ! 2 ! 2 !}}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}{}}$
We can write the above equation as
$\begin{array}{l} =\frac{10 ! 4 !}{3 ! 4 ! 2 ! 2 !} \times \frac{3 ! 4 ! 2 ! 2 !}{13 !} \\ =\frac{10 ! \times 4 !}{13 \times 12 \times 11 \times 10 !} \\ =\frac{4 \times 3 \times 2 \times 1}{13 \times 12 \times 11} \end{array}$
$=\frac{2}{143}$