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

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In how many ways can the word "illusion" be rearranges such that the t  [#permalink]

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New post 05 Feb 2018, 00:19
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

75% (01:23) correct 25% (01:28) wrong based on 56 sessions

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Re: In how many ways can the word "illusion" be rearranges such that the t  [#permalink]

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New post 05 Feb 2018, 00:49
Bunuel wrote:
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!


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.
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Re: In how many ways can the word "illusion" be rearranges such that the t  [#permalink]

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New post 05 Feb 2018, 00:51
Bunuel wrote:
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!


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)
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Re: In how many ways can the word "illusion" be rearranges such that the t  [#permalink]

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New post 11 Feb 2018, 11:29
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!
Re: In how many ways can the word "illusion" be rearranges such that the t &nbs [#permalink] 11 Feb 2018, 11:29
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