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word formation... [#permalink]

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New post 21 Jan 2006, 11:54
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I know it must be very easy for most of you, so please explain me:
how many different words with 3 letters can we make using the letters:
R,R,R,F,F,T,T ?

If I count them they seem to be 19, but I would like to know a faster way to find it...

RRR
FFT
FFR
TFF
RFF
FTF
FRF
RRT
RRF
TRR
FRR
RTR
RFR
TTR
TTF
RTT
FTT
TFT
TRT
---------------=19
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New post 21 Jan 2006, 12:08
Well here is what I thought:

Total words = Words with 3 letters repeted + No letters repeted + Words 2 letters repeted

Words with 3 letter repetitions = 1 = RRR
Words with No letters repeted (R,T,F) = 3! = 6

Words with 2 letetrs repeated:
2Rs: 2R's,1F or 2R's,1T = 3*2 =6
Similarly for 2Fs = 6

Total = 1 + 6 + 6 *2 = 19
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New post 21 Jan 2006, 18:54
thank you for your answer giddi77!
I can understand your reasoning except from this part:

"Words with 2 letetrs repeated:
2Rs: 2R's,1F or 2R's,1T = 3*2 =6
Similarly for 2Fs = 6 "

You covered 2Rs and 2Fs but where are the 2Ts ?
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New post 21 Jan 2006, 21:26
eleni24 wrote:
thank you for your answer giddi77!
I can understand your reasoning except from this part:

"Words with 2 letetrs repeated:
2Rs: 2R's,1F or 2R's,1T = 3*2 =6
Similarly for 2Fs = 6 "

You covered 2Rs and 2Fs but where are the 2Ts ?


Oops! I somehow assumed that there is only 1 T:

Infact for the 2T's there will be 6 more combinations:
2T's,1F or 2T's,1R
2T's,1F = 3!/2! (3 combinations in where 2 letters are repeated) = 3
2T's,1R = 3!/2! = 3

So final answer = 1 + 6 + 6*3 = 25!

What you are missing in your list are single letter combinations:
RTF
RFT
TRF
TFR
FTR
FRT

So the total would be 25! HTH.
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New post 22 Jan 2006, 00:34
Can anyone explain why this formulae doesnt work here

7 letters with 3R's 2F's and 2T's

so 7P3/3!2!2!
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New post 22 Jan 2006, 13:07
andy_gr8 wrote:
Can anyone explain why this formulae doesnt work here

7 letters with 3R's 2F's and 2T's

so 7P3/3!2!2!


Andy I am not sure how 7P3/3!2!2! is correct, infact it doesn't even result in a n integer value! Moreover why should we divide the number by (3!*2!*2!) if in every word the we consider there wont be all the possible characters.

If all the letters are considered then yes it would be 7!/3!2!2!
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New post 22 Jan 2006, 21:54
Agree with giddi77 except for the factorial notation :).


giddi77 wrote:
eleni24 wrote:
thank you for your answer giddi77!
I can understand your reasoning except from this par:

"Words with 2 letetrs repeated:
2Rs: 2R's,1F or 2R's,1T = 3*2 =6
Similarly for 2Fs = 6 "

You covered 2Rs and 2Fs but where are the 2Ts ?


Oops! I somehow assumed that there is only 1 T:

Infact for the 2T's there will be 6 more combinations:
2T's,1F or 2T's,1R
2T's,1F = 3!/2! (3 combinations in where 2 letters are repeated) = 3
2T's,1R = 3!/2! = 3

So final answer = 1 + 6 + 6*3 = 25!

What you are missing in your list are single letter combinations:
RTF
RFT
TRF
TFR
FTR
FRT

So the total would be 25! HTH.
  [#permalink] 22 Jan 2006, 21:54
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