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# word formation...

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21 Jan 2006, 10:54
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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21 Jan 2006, 11:08
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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21 Jan 2006, 17:54
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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21 Jan 2006, 20:26
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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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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21 Jan 2006, 23:34
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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22 Jan 2006, 12:07
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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22 Jan 2006, 20:54
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.
22 Jan 2006, 20:54
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