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March 8th, 2015, 06:13 AM   #1
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Combinatorics. How to select spaces for letters.

Hi,

I have the following problem:
In how many ways can you arrange the letters of the word OBLIGE, without letting the letters B,E be together?


so, I take all the letters except B,E, putting one space between them where I can insert the two letters I said before. See the following schema:

_ O _ L _ I _ G _

I) here, each time I select one letter, then I go to select one space that will host this last letter, so:
I go to choose 1 out of 2 letters {B,E} and also 1 out of 5 spaces for the letter just selected, and, I go to choose 1 out of 1 remaining letter and 1 out of 4 remaining spaces for the letter just selected:
$\displaystyle \binom{2}{1} \cdot \binom{5}{1} \cdot \binom{1}{1} \cdot \binom{4}{1} = (2 \cdot 5) \cdot (1 \cdot 4) = 10 \cdot 4 = 40$
but, this result is not correct.


the right reasoning would be the following:
II) At first, I consider every necessary space to insert the letters I want, and then I go to select the wanted letters:
$\displaystyle \binom{2}{1} \cdot \binom{1}{1} \cdot \binom{5}{2} = 2 \cdot 1 \cdot 10 = 20$
identical result of: $\displaystyle {}^5P_{2} = 5 \cdot 4 = 20$

the letters OLIG can be arranged in 4!, so, the number of ways to arrange the entire word is: $\displaystyle 4! \cdot 20 = 480$


my question is:
even if both reasonings seems to me logically right, please, can you explain me, why it's wrong the reasoning I), but it's right the reasoning II) ?

many thanks!
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March 9th, 2015, 12:24 PM   #2
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Hello, beesee!

Quote:
In how many ways can you arrange the letters of
the word OBLIGE, without letting the letters B,E be together?

I can't follow your first solution.


_ O _ L _ I _ G _








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