# Arrangement of letters in a word

1. Jul 9, 2015

### rajeshmarndi

1. The problem statement, all variables and given/known data
In how many ways 4-lettered words that can be formed using the letters of the word "BOOKLET".

The book solution solve it this way.
The word contains 7 letters out of which there are 2 O's. So there are 6 letters.
∴ The number of 4 lettered words
= 7P4 - 6P4 = 840 - 360 = 480

2. Relevant equations

3. The attempt at a solution
What we need is to deduct the duplicate counting of 2 O's from the total permutation.
So how does 6P4 solve this purpose.

isn't it should have been
( 4P2 * 5P2 ) / 2
i.e number of ways these 2 O's can be placed in 4 place * number of ways B,K,L,E,T can be placed in the remaining 2 places divide by 2.

= (12 * 20)/2 =120

And the answer would be 840 - 120 = 720 and not 480.

2. Jul 9, 2015

### Orodruin

Staff Emeritus
It is not only the words with two Os that are double counted in 7P4.

3. Jul 9, 2015

### rajeshmarndi

Correction it is 6P4

But I do not see any other letters being doubled.
My question, is the book solution, correct, any way?

4. Jul 9, 2015

### haruspex

Imagine the two Os in different colours. In all the words corresponding to 7P4, which words only differ in colour pattern?
(I agree with the book answer, though I've yet to understand how they obtain it by subtracting 6P4.)

5. Jul 10, 2015

### rajeshmarndi

I understood that and that's why I mentioned them as O1 & O2.
I am not sure if I've understood you correctly, if we imagine the two Os in different colours(and each unique letter to be a different colour), then 7P4 will result all to be in different colour.[/QUOTE]
Where am I wrong in my below solution.

6. Jul 10, 2015

### haruspex

Consider BLO1K, BLO2K.

7. Jul 10, 2015

### Orodruin

Staff Emeritus
I called them O and O' in my mid, the principle is the same. I too have not understood how they arrive at it being 6P4 other than numerical coincidence. I solved it by first considering all words as double counted and then adding those that this removes by error (i.e., half of the 4 letter words made from BKLET).

8. Jul 10, 2015

### rajeshmarndi

Yes, thanks, it just didn't hit my mind.

9. Jul 10, 2015

### rajeshmarndi

But, I am still not yet clear.

As 7P4 - 6P4 = number of words containing 2 Os (including O1 & O2)

, since 6P4 = total permutation without 2 Os in the word.

10. Jul 10, 2015

### haruspex

Yes, but it's coincidence. 360 have no Os, 240 have one O, 180 have two. It just happens that 360=(240+180)/2.
I can make an argument for subtracting 46P3. Consider the number that have a specific O (whether or not they have both).

11. Jul 10, 2015

### Orodruin

Staff Emeritus
5! = 120 has no Os, 4*5*4*3 = 240 has one O, (4!/(2!)^2)*5*4 = 120 has two Os. The sum of all of these is 480, alternatively you subtract 240+120 = 360 from 840 and get the same result.

Average of 240 and 180 is 360?

12. Jul 10, 2015

### haruspex

I shouldn't post while watching Le Tour.