Question

If all the letters in the word “SUCCESS” are written down at random in a row, then the probability that no two C’s and no two S’s occur together.
A. $\dfrac{2}{35}$
B. $\dfrac{8}{35}$
C. $\dfrac{2}{7}$
D. None of these

Answer 1

We will first find the total number of possible ways to write the word “SUCCESS” and we consider it as the total number of events in the sample space. For finding the probability of an event we need the total number of favorable cases for the event in the sample space. For that we assume the word without S’s and find the number of possible ways to write the word without two S’s come together. After that we will find the number of possible ways to write the word without two S’s and two C’s come together. Finally we will find the probability by taking the ratio of number of possible ways to the total number of events in the sample space.

Complete step by step answer:
Given that,
The word “SUCCESS” has following number of letters
$\begin{aligned} & S\to 3 \\\ & U\to 1 \\\ & C\to 2 \\\ & E\to 1 \\\ \end{aligned}$
The word has $7$ letters out of them $3$ are S’s and $2$ are C’s. Hence the number of ways to write the word “SUCCESS” is
$\begin{aligned} & n\left( s \right)=\dfrac{7!}{3!\times 2!} \\\ & =420 \end{aligned}$
Now we have to find the number of ways to write the word without no two S’s come together then we must write the word as \\_\text{U }\\!\\!\\_\\!\\!\text{ C }\\!\\!\\_\\!\\!\text{ C }\\!\\!\\_\\!\\!\text{ E }\\!\\!\\_\\!\\!\text{ }. In the $5$ blanks we can write $3$ S’s in ${}^{5}{{C}_{3}}$ ways and the remaining $4$ letters are arranged in $\dfrac{4!}{2!}$ ways due to repetition C’s. So, we can write the word without two S’s come together in ${}^{5}{{C}_{3}}\times \dfrac{4!}{2!}=120$ ways.

But here we have a chance to get two C’s come together. To avoid this, we can write the word as \\_\text{U }\\!\\!\\_\\!\\!\text{ CC }\\!\\!\\_\\!\\!\text{ E }\\!\\!\\_\\!\\!\text{ }. Here we can fill the $4$ blank with $3$S’s in ${}^{4}{{C}_{3}}$ ways and the remaining $3$ places are filled with remaining word in $3!$ ways. So, we can form the word \\_\text{U }\\!\\!\\_\\!\\!\text{ CC }\\!\\!\\_\\!\\!\text{ E }\\!\\!\\_\\!\\!\text{ } in ${}^{4}{{C}_{3}}\times 3!=24$ ways.

So, the number of ways to write the given word where no two C’s and no two S’s occur together is $120-24=96$.
Hence the probability is $\dfrac{96}{420}=\dfrac{8}{35}$

So, the correct answer is “Option B”.

Note: Please make sure that we have found all the possible ways to make the word with the given condition and we have asked to find the probability of the event don’t stop your solution after finding the number of possible cases for the given event.