In how many ways can the word "illusion" be rearranges such that the t

Question

In how many ways can the word "ILLUSION" be rearranges such that the two l are NOT together?

A. 8! − 8
B. 7*7!
C. 7!*2!*5!
D. 8!/(2!*2!) - 7!/(2!)
E. 6!

pushpitkc · Accepted Answer

The total number of ways in which ILLUSION can be arranged is  \frac{8!}{(2!)(2!)}

We have been asked to find the number of ways in which the word ILLUSION can be 
re-arranged such that the two II's never occur together. Lets consider the two IIs to 
be an X - so the ways XLLUSON can be re-arranged are  \frac{7!}{(2!)}

Therefore, the number of ways the word can be arranged is  \frac{8!}{(2!)(2!)} - \frac{7!}{(2!)}  (Option D)

DavidTutorexamPAL · Answer

The difficulty in combinatorics questions is often in figuring out what you need to do.
We'll look for the right choice model, a Logical approach.

We'll start with what we know, that the two 'l' cannot be together.
There are 8C2 ways to pick two locations in an 8-letter word and we can subtract from this 7 adjacent and therefore impossible locations.
Within these 8C2 - 7 arrangements we need to choose the order of the letters (which l comes first) but because they are both identical then it doesn't matter which comes first.
(If they were different letters we would have needed to multiply by 2!)
Now we need to arrange all the other 6 letters giving 6! options and divide by 2! because there are two identical I's.

So, we have a total of (8C2 - 7)6!/2! options.
This simplifies to (8!6!)/(6!2!2!) - (7*6!)/2! = 8!/(2!2!) - 7!/2!

(D) is our answer.

beesh1 · Answer

1. Possible illusion order:
(8)!/(2!2!)

2. Ways Illusion can be ordered IF the two I's are together: 
Place the two I's together, and count them as 1.
X L L U S O N
Total letters now: 7
Repeating letters: L
This becomes: 7!/2!

(8)!/(2!2!) - 7!/2!

kapstone1996 · Answer

Am I correct to say the reason for the 8!/2!2! is because of the two "I"s and two "l"s?

Then since we do not want the two "I" together that we get 7!/2!, where again the 2! in the second equation represents the two "l"s

Which we end up with 8!/2!2! - 7!/2!

I chose D, but was an educated guess.

ScottTargetTestPrep · Answer

The number of ways to rearrange ILLUSION when the I’s are together is.

-L-L-U-S-O-N

We have 7 total “spots” and   can be arranged in just one way. Also, we must divide by 2! to account for the two (indistinguishable) L’s.

Thus, the number of arrangements is 7!/2!

The total number of ways to arrange the letters of I-I-L-L-U-S-O-N is:

8!/(2! x 2!)

Thus, the number of ways the word "ILLUSION" can be arranged such that the two l’s are NOT together is 8!/(2! x 2!) - 7!/2!

Answer: D

Aliprep · Answer

Why is different combinations for word ILLUSION is 8!/2!2!? (not the question that is asked, but the first part of the calculations, that everyone calculates?)
Isn't it 8!?

Bunuel · Answer

THEORY 
 
Permutations of  n  things of which  P_1  are alike of one kind,  P_2  are alike of second kind,  P_3  are alike of third kind ...  P_r  are alike of  r_{th}  kind such that:  P_1+P_2+P_3+..+P_r=n  is:
  \frac{n!}{P_1!*P_2!*P_3!*...*P_r!} 
 
For example the number of permutation of the letters of the word "gmatclub" is  8!  as there are 8 DISTINCT letters in this word.
 
The number of permutation of the letters of the word "google" is  \frac{6!}{2!2!} , as there are 6 letters out of which "g" and "o" are represented twice.
 
The number of permutations of 9 balls out of which 4 are red, 3 green and 2 blue, would be 9!/4!3!2!.

So, the number of permutation of the letters of the word "ILLUSION" is  \frac{8!}{2!2!} , as there are 8 letters out of which "L" and "I" are represented twice.

21. Combinatorics/Counting Methods 
     Theory   
 Math: Combinatorics  
 Combinatorics Made Easy! 
 1 hour Combinatorics Video Lesson 
 Combinations while dealing with similar and dissimilar things 
 When Permutations & Combinations and Data Sufficiency Come Together 
     Questions   
 Advanced Topic: Probability and Combinations Questions With Solutions 
 Letter Arrangements in Words 
 Seating Arrangements in a Row and around a Table 
 Constructing Numbers, Codes and Passwords 
 DS Questions 
 PS Questions

For more:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

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In how many ways can the word "illusion" be rearranges such that the t

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