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# The letters "CFOSU" are arranged in dictionary order. What

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The letters "CFOSU" are arranged in dictionary order. What [#permalink]

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19 Apr 2005, 17:38
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The letters "CFOSU" are arranged in dictionary order. What is the
rank of the word "FOCUS" in this order?

(a) 17
(b) 21
(c) 31
(d) 35
(e) 38
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19 Apr 2005, 23:44
31...

C in first place = 4! ways

F in first place and C in second place = 3! ways

F in 1st p, O in 2nd p and C in 3rd p = 2 ways => FOCSU and then FOCUS

4!+3!+1=31
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20 Apr 2005, 02:42
christoph wrote:
31...

C in first place = 4! ways

F in first place and C in second place = 3! ways

F in 1st p, O in 2nd p and C in 3rd p = 2 ways => FOCSU and then FOCUS

4!+3!+1=31

There are 31 words before FOCUS. Therefore shouldnt FOCUS' rank be 32nd? There's no such option though
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20 Apr 2005, 05:00
kapslock wrote:
christoph wrote:
31...

C in first place = 4! ways

F in first place and C in second place = 3! ways

F in 1st p, O in 2nd p and C in 3rd p = 2 ways => FOCSU and then FOCUS

4!+3!+1=31

There are 31 words before FOCUS. Therefore shouldnt FOCUS' rank be 32nd? There's no such option though

combinations before FOCSU are 4!+3!...FOCSU is 31...and FOCUS is 32
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20 Apr 2005, 16:17
A longer but elaborate method: (From Math Forum)

Let's use a code to solve this problem. We have the following letters:

C

F

O

S

U

And we will take all the permutations of these letters and arrange
them in alphabetical order.

I think it will be helpful to think of these letters as numbers. The
first letter, alphabetically, in this list is C, so let's replace "C"
with the number 1.

The second letter is F, so let's replace it with 2.

And so on.

Using this code, the word "FOCUS" will be the sequence:

2 3 1 5 4

The first permutation in the sequence "alphabetically" is:

1 2 3 4 5

Then

1 2 3 5 4

Then

1 2 4 3 5

And it goes on like that.

The permutation we want is, again:

2 3 1 5 4

Which starts with a 2, so it comes AFTER every permutation that starts
the number of ways the digits:

2 3 4 5

can be arranged.
tells us that that number of combinations is going to be 4! (4*3*2*1)
or 24.

So there are 24 permutations that begin with 1. That means that the
25th permutation is:

2 1 3 4 5

Now we've gotten the first digit in place. The next digit we want is
3. So we are going to have to work through all of the permutations
that start with "2 1". How many are there? 3! (3*2*1) or 6.

So the 30th permutation is:

2 3 1 4 5

And the 31st permutation is:

2 3 1 5 4
Solution   [#permalink] 20 Apr 2005, 16:17
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