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ps combinatorics

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ps combinatorics [#permalink] New post 01 Sep 2004, 11:15
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A
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D
E

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6 ppl need to divide into 3 groups of 2

is it 6c2 * 4c2 * 2c2*??

another one... 5 letter word

DEFGG

how mnay words can you make where the two G's are not together!
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 [#permalink] New post 01 Sep 2004, 11:29
Your first question needs to be divided by 3! because order of group does not count. Hence, answer should be 6c2*4c2/3!

Second question is 5! - 4! = 120 - 24 = 96
4! is total number of unfavorable outcomes when 2 Gs(as a single entity) are next to each other:
(GG) - D - E - F --> 4! ways of sorting these differently
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 [#permalink] New post 01 Sep 2004, 11:30
You are right on the first combination equation.

As for the word, my guess is that there are 5! total words that can be made....120. There are 5C2 unfavorable combinations of having the two G's together. So 120-5C2=110.
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 [#permalink] New post 01 Sep 2004, 11:39
I am wrong for the second question:
Total outcome should be 5!/2! because there are 2 G's which can be interchanged. --> 60
4! would be all unfavorable outcomes as previously explained
Answer should be: 60 - 24 = 36
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 [#permalink] New post 01 Sep 2004, 11:41
I would like to underline one point and gather your comments guys :

About the 2nd question, the 96 answer considers that all words are 5 letters long and the two G are distinct.
When I first tried to answer I was thinking about 5 ar less characters words...
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 [#permalink] New post 01 Sep 2004, 12:54
thanks guys

36 is the caorrect answer for the second problem
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 [#permalink] New post 01 Sep 2004, 19:05
6 ppl need to divide into 3 groups of 2

is it 6c2 * 4c2 * 2c2*??

Hah ! Let me try something Ian taught me yesterday and see if I get this right !

6C2*4C2*2C2/3! = 15*6*1/6 = 15 ways
-----------------------------------------------------------------

another one... 5 letter word

DEFGG

how mnay words can you make where the two G's are not together!

Do the words have to be real words ? Or just any combination of letters as long as the two Gs are not together.

If the words have to be real:
Then the number of words = 0 (The only word that can be made is egg, and it's out because the two g's are together)

If it's any letters and the words do not need to make sense:
Then , we can either has 1 letter word, 2 letter words, 3 letter words, 4 letter words, can't be 5 as that will give 2 g's
So 1 letter word - 5
2 letter words - 4P2 (remove 1 G) = 12 (use permutation assuming we can swtich the letters around and form another word)
3 letter words - 4P3 = 24
4 letter words - 4P4 = 16

So total - 4 + 12 + 24 + 16 = 56
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 [#permalink] New post 02 Sep 2004, 08:18
ywilfred, I believe that you have to use all the letters to form a 5 letter word. Also, why did you omit to account for 5 letter words in your calculation? Although there are 2 G's, you should still account for words that can be formed when the 2 G's are apart.
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 [#permalink] New post 03 Sep 2004, 23:56
The answer to the first problem according to my working is
6C2*4C2*2C2/3!
=15*6/6=15

In the word problem, the number of different words possible is 5!/2! ( as G is coming twice)
=60

If G and G is treated as one unit, number of ways that the rest of the letters can be arranged= 4!=24

The number of ways in which the two Gs will be apart
=60-24
=36
  [#permalink] 03 Sep 2004, 23:56
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