Question

The number of ways of arranging the letters of the word SUCCESSFUL so that all s’s will not come together is:
A. $\dfrac{10!}{3!2!2!}-\dfrac{8!}{2!2!}$
B. $\dfrac{9!}{3!2!3!}-\dfrac{8!}{2!2!}$
C. $\dfrac{9!}{2!2!}-\dfrac{8!}{3!}$
D. $\dfrac{10!}{8!2!}$

Answer 1

Hint: To solve this question, we should use the concept of combinations. The number of ways of arranging n objects in which there are p are identical of one category, q are identical of another category and r are identical of another is given by $\text{Ways}=\dfrac{n!}{p!q!r!}$. Using this formula, we can write the total number of ways in which the letters of the word SUCCESSFUL can be arranged. To get the ways in which all s’s are not together, we should subtract the number of such ways from the total number of ways. We should assume that the letter SSS as an object. Then, we will have 8 objects and we will use the same formula to get to the answer.

Complete step-by-step answer:
The number of ways of arranging n objects in which there are p are identical of one category, q are identical of another category and r are identical of another is given by $\text{Ways}=\dfrac{n!}{p!q!r!}$.
Let us consider the word SUCCESSFUL. We have
3 – S, 2 – U, 2 – C, 1 – E, 1 – F, 1 – L.
The total number of letters is 10. So, we can write that
$n=10$, $p=3,q=2,r=2$. We can write the total number of ways as
$\text{Total Ways}=\dfrac{10!}{3!2!2!}$

Now, to get the required answer, we will subtract the number of ways such that all the s’s are together from the total ways.
Let us consider that s’s are together. It means that we should assume SSS as a single object. We have
1 – SSS, 2 – U, 2 – C, 1 – E, 1 – F, 1 - L.
Total objects are 8 in number. We can write the number of ways as
$\text{Ways}=\dfrac{8!}{2!2!}$
The required ways are the difference of the total ways and ways with all s’s together. So, we can write that
$\text{Required ways}=\text{Total Ways }-\text{ Ways with all 3s }\!\!'\!\!\text{ s together}$
$\text{Required Ways}=\dfrac{10!}{3!2!2!}-\dfrac{8!}{2!2!}$
$\therefore $The required number of ways is $\text{Required Ways}=\dfrac{10!}{3!2!2!}-\dfrac{8!}{2!2!}$.

So, the correct answer is “Option A”.

Note: Some students might misinterpret what is asked in the question. They think that s’s should not be consecutive to one another but we are actually asked the cases in which all the three s’s are not together. So, there can be two s’s consecutive to each other in our question. Students’ misinterpretation will lead to a different answer. So, in these types of problems in permutations and combinations, understanding each and every word of the question is important.