Question

If the letters of the word ASSASSIN are written in a row at random, the probability that no two S’s come together is
(a) $\dfrac{1}{7}$
(b) $\dfrac{2}{7}$
(c) $\dfrac{1}{14}$
(d) $\dfrac{3}{14}$

Answer 1

Find the total number of words that can be formed with the letters of the word ASSASSIN. Use the relation: - $\dfrac{n!}{m!}$, where ‘n’ is the number of letters and ‘m’ is the number of times any letter repeats. Now, fill the letters A, A, I and N at four places and fill the four S’s such that no two of them occur together. To do this, keep A, A, I and N between two S’s. Find such a number of possible words and take it’s ratio with the total number of words formed to get the probability.

Complete step by step answer:
Here, we have been provided with the words ASSASSIN. There are 8 letters in which ‘S’ is occurring 4 times and ‘A’ is occurring two times. So, the total number of words that can be formed by using the given 8 letters = $\dfrac{8!}{4!\times 2!}$, because there are 4 S’s and 2 A’s.
$\Rightarrow$ Total number of words that can be formed = $\dfrac{8!}{4!\times 2!}$.
Now, we can form such words where two S’s cannot occur together. So, let us arrange A, A, I and N at four places.
1 A 2 A 3 I 4 N 5
Here, there are 5 vacant places where four S’s are to be filled. Clearly, we can see that in any way we will fill these four S’s there will always be a different letter between two S’s. So, two S’s will never occur together. Therefore,
Number of ways to select four vacant places out of five = ${}^{5}{{C}_{4}}=5$
Now, A, A, I and N can be arranged among themselves.
$\Rightarrow$ Number of ways to arrange A, A, I and N =$\dfrac{4!}{2!}$, since A is occurring twice.
So, the total number of arrangements such that no two S’s come together = $5\times \dfrac{4!}{2!}$.
Therefore, the required probability will be the ratio of $\dfrac{5\times 4!}{2!}$ and $\dfrac{8!}{4!\times 2!}$.
$\Rightarrow$ Probability = $\dfrac{5\times 4!}{2!\times \dfrac{8!}{4!\times 2!}}$
$\Rightarrow$ Probability = $\dfrac{5\times 4!\times 4!\times 2!}{2!\times 8!}$
On simplifying the above expression, we get,
Probability = $\dfrac{1}{14}$

So, the correct answer is “Option C”.

Note: One may note that there are five vacant places in which four S’s can be filled. So, we had to select only four out of these five places. That is why we have used the expression ${}^{5}{{C}_{4}}$. Always remember that if a letter is repeating ‘n’ times then we have to divide the total arrangement possible with n!.