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I am facing difficulty in solving the 18th Question from the above passage.


I have attempted it using principal of inclusion and exclusion.

Let n be the required number of ways.

$n = \text{Total number of ways - Number of ways in which Os and Is are together}$

$\implies n = \dfrac{12!}{3!2!2!} - \dfrac{11!}{2!3!} - \dfrac{10!}{2!2!}+ \dfrac{9!2!}{2!} = 399 \times 8!$

This gives $N = 399$ which is wrong. Can you please tell me why am I getting the wrong answer?

  • 1
    $\begingroup$ How do you know that your solution is wrong? There is also option (D) $\endgroup$ – Robert Z Mar 28 at 8:29
  • $\begingroup$ @RobertZ answer given is option B $\endgroup$ – Abcd Mar 28 at 8:30

Your mistake is not noticing that the I's can be non-separated two different ways: We can have all three of them together as III, but we can also have II in one place and I in another.

One way to calculate how many ways this can happen is to think of one pair II as a single letter, and the third I as a separate letter, and count the number of words we can make. This will count each III case twice: Once as I II, and once as II I. So we have to subtract the number of such cases once.

With this correction, the final calculation becomes $$ \overbrace{\frac{12!}{3!2!2!}}^{\text{All words}} - \overbrace{\frac{11!}{3!2!}}^{\text{O's not separate}} - \overbrace{\left(\underbrace{\frac{11!}{2!2!}}_{\text{II and I}} - \underbrace{\frac{10!}{2!2!}}_{\text{III}}\right)}^{\text{I's not separate}} + \overbrace{\left(\underbrace{\frac{10!}{2!}}_{\text{OO, II and I}} - \underbrace{\frac{9!}{2!}}_{\text{OO and III}}\right)}^{\text{Neither O's nor I's separate}} $$ which turns out to be $8!\cdot 228$.

  • $\begingroup$ Thanks for the answer I have understood my mistake, but I am unable to understand the final equation in your answer. Can you label the terms like which term stands for which case? $\endgroup$ – Abcd Mar 28 at 9:21
  • $\begingroup$ @Abcd I added a short explanation for what each term represents. The basic structure of your inclusion-exclusion calculation is still there, just with the two last terms corrected to make up for the mistake. $\endgroup$ – Arthur Mar 28 at 9:32
  • 1
    $\begingroup$ +1 for the MathJax magic! $\endgroup$ – TonyK Mar 28 at 12:09

The letter $I$ appears three times. You have calculated the number of ways to arrange the letters so that the three copies of $I$ aren't all in the same place (and also the two $O$s are separated). However, "separated" here means that no two of them are together. Your count includes such "words" as STIILOCATINO, which should be excluded.


How did you get each of the terms of $n$?

  • $\frac{12!}{3!2!2!}$ : The number of ways of arranging the letters of SOLICITATION.

  • $\frac{11!}{2!3!}$ : The number of ways in which the two $O$s can be together in SOLICITATION.

  • $\frac{10!}{2!2!}$ : The number of ways in which the three $I$s can be together in SOLICITATION.

  • $\frac{9!2!}{2!}$ : The number of ways in which the three $I$s and two $O$s can be together in SOLICITATION.

Well , the problem is this : the complement of this case, also includes the case where two $I$s can be together and the third can be separate. You have not neglected this case, and that's why you have an answer larger than the given answer.


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