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word formation...

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eleni24
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word formation... [#permalink] New post   21 Jan 2006, 11:54
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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giddi77
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Re: word formation... [#permalink] 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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eleni24
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Re: word formation... [#permalink] 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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giddi77
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Re: word formation... [#permalink] 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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Re: word formation... [#permalink] 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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giddi77
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Re: word formation... [#permalink] 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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Re: word formation... [#permalink] 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.




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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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Re: word formation... [#permalink]
22 Jan 2006, 21:54
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