My Math Forum Combinatorics. How to select spaces for letters.

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 March 8th, 2015, 06:13 AM #1 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 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!
March 9th, 2015, 12:24 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Hello, beesee!

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

$\quad$ _ O _ L _ I _ G _

$\text{There are }4!\,=\,24\text{ choices to arrange the OLIG.}$

$\text{Then we select two of the five spaces for the B and E:}$
$\;\;\;\; _{^5}P_{_2}= 20\text{ ways.}$

$\text{Therefore, there are: }\:24\,\cdot\,20 \:=\:480\text{ ways.}$

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