My Math Forum counting problem

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 February 20th, 2012, 11:48 AM #1 Newbie   Joined: Dec 2011 Posts: 15 Thanks: 0 counting problem Hi, How many different ways can you arrange the letters of the word "VISITING" such that the letters ing (in all 3! possible cases) are next to each other.
February 20th, 2012, 01:43 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
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Re: counting problem

Hello, scream!

I think I've solved it . . .

Quote:
 How many different ways can you arrange the letters of the word "VISITING" such that the letters ING (in all 3! possible cases) are next to each other?

$\text{Duct-tape an I, an N and a G together.}
\text{\;\;\;\;\;Note that they can have }3!\text{ possible orders.}
\text{Then we have six "letters" to arrange: }\:\fbox{ING},\; I,\;I,\;S,\;T,\;V.$

$\text{If we had six }different\text{ letters, there would be }6!\text{ permutations.}
\text{\;\;\;\;\;And the number of ways would be: }\:3!\,\cdot\,6!
\text{But we have 3 identical I's which can be switched in }3!\text{ ways with no effect.}$

$\text{Therefore, there are: }\:\frac{3!\,\cdot\,6!}{3!} \:=\:720\text{ ways.}$

 February 21st, 2012, 11:50 AM #3 Newbie   Joined: Dec 2011 Posts: 15 Thanks: 0 Re: counting problem Thanks a lot soroban. I think 6! is not true. Consider the smaller problem for letters of the word "VIIIN" such that IN are next to each other. With your reasoning the number of all possible words should be (4!)(2!)/(3!)=8, but the answer is 18. I think I have found true answer of the original problem. It is 6!-2.5.5!=1920, but I am not sure yet. Thanks again.

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