Find the number of permutations of the letters of the word ”ILLINOIS” for which ”I” does not occur at the left of ”L”

Question

where 'ILONSLII' is not accepted , but 'INLOISLI' it's accepted

this is my answer have i mistake ?

we will divided our problem into 3 separate cases

• if LL are consecutive and in the first position , here there are $\frac{6!}{3!}=120$ permutations for the remains letters "IIINOS"
• if LL are consecutive and not in the first position , here we will consider "LL" as a one letter so there are 6 position of "LL" and 3 ways for the letters before "LL" which are "NOS" not "I" , and $\frac{5!}{3!}$ ways for "III and 2 letters that not placed before LL", so there are $6\times3\times\frac{5!}{3!}=360$

  • if "LL" aren't consecutive , there are $\frac{6!}{3!}$ ways for "IIINOS" without the letter "L" , and for "L" $\binom{4}{2}$ ways , which are $\bigodot I \bigotimes I\bigotimes I\bigotimes N\bigodot O\bigodot S\bigodot$ then there are $\frac{6!}{3!} \times \binom{4}{2}=6!$ ways

    Hence,there are $120+360+6!$ ways

Answer 1

Your answer is correct, I haven't checked the process as I'm suggesting a much simpler process

• Taking all letters to be distinct initially, permute the $5$ letters excluding the $I's$ in $5!$ ways

• Whichever such arrangement you take, there are $6$ gaps between letters (including ends) where the first $I$ could be put, but two spots immediately to the left of the two $L's$ are barred, so $4$ valid spots to put the first $I$

• Each $I$ put creates an extra spot for the next $I$, so the three $I's$ can be placed in $4\cdot5\cdot6$ ways

• Putting it all together, and correcting for repeated letters, valid arrangements $= \dfrac{5!\cdot4\cdot5\cdot6}{3!2!} = 1200$

Answer 2

Alternative approach is Inclusion-Exclusion.

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

Following the syntax in the second link:

• Let $~S~$ denote the set of all of the ways that the letters in ILLINOIS can be permuted.

• For $~k \in \{1,2,\cdots,7\},~$ let $~S_k~$ denote the subset of $~S~$ where there is an I in position $~k,~$ and an L in position $~(k+1).~$ Here, I am counting the positions, from the left. Note that (for example), for a random element in $~S_1,~$ there may or may not be a second occurrence of IL, after position 2.

Then, the desired computation is

$$|S| - |S_1 \cup \cdots \cup S_7|.$$

The Inclusion-Exclusion approach is much easier here, than it would seem, because:

• There are shortcuts because of considerations of symmetry.

• Since there are only 2 L's in ILLINOIS, any intersection of (for example) $~S_{i_1} \cap S_{i_2} \cap S_{i_3}~$ must be the empty set.

So, the final computation will be $~T_0 + T_2 - T_1,~$ where

• $T_0 = \displaystyle \binom{8}{3} \times \binom{5}{2} \times 3! = 3360.$

• $T_1 = \displaystyle \binom{7}{1} \times \binom{6}{2} \times 4! = 2520.$

So, the entire problem reduces to computing $~T_2.~$ This computation has a minor complication. The assumption that the first factor in the computation is $~\displaystyle \frac{\binom{7}{1} \times \binom{5}{1}}{2}~$ is wrong. In fact, this (wrong) factor isn't even an integer. The inaccuracy in this factor is illustrated by the following analysis:

• If the left-most set being intersected is $~S_1,~$ then there are $~5~$ choices for the second set $~S_k.~$ These choices are $~k \in \{3,4,\cdots,7\}.~$

• If the left-most set being intersected is $~S_2,~$ then there are only $~4~$ choices for the second set $~S_k.~$ These choices are $~k \in \{4,5,\cdots,7\}.$
  
  The (hidden) complication is that if $~S_2~$ is one of the sets being intersected, then there can be no intersection that starts in position 1.

• So, the left-most set being intersected can start at position $~k,~$ where $~k \in \{1,2,\cdots,5\},~$ and for each value of $~k,~$ there are $~(6 - k)~$ positions available for the right-most set being intersected.
  
  Therefore, the first factor in the computation of $~T_2~$ will be $~\displaystyle \sum_{k=1}^5 (6 - k) = 15.$

Therefore,

$$T_2 = 15 \times 4! = 360.$$

Therefore, the final computation is

$$(T_0 + T_2) - T_1$$

$$=(3360 + 360) - 2520 = 1200.$$