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Permutation and Combination

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Intern
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Permutation and Combination [#permalink] New post 19 Sep 2005, 22:29
Hi

Help me with the P&C question below, tell simple way. Tell me links where I can find more of P&C stuff.

In how many ways can the letters of the word ABACUS be
rearranged such that the vowels always appear together?

A. 6!/2!
B. 3!*3!
C. 4!/2!
D. 4! *3!/2!
E. 3!*3!/2

Thanks
Director
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 [#permalink] New post 20 Sep 2005, 03:26
ABACUS has 6 letters, 3 vowels and 3 consonants .Take all the vowels as one unit then u have 4 elements- each of the consonants and a group of 3 vowels. This can be ordered in 4! ways. But don't forget that the group of 3 vowels can have 3! orderings within the group. Or we have 4!x3! and finally cause u have 2-A's you need to divide the whole product by2!, or your ans should be (4!x3!)/2! or option D)
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 [#permalink] New post 20 Sep 2005, 06:11
Same here, intuitively. 72 is the answer.

There are 4 places where the string of 3 vowels can appear:
VVVCCC
CVVVCC
CCVVVC
CCCVVV

The string of vowels can be arranged in 3 ways (because two of the letters are identical As):
AAU
AUA
UAA

The string of consonants can be arranged in 6 ways (because the letters are all different):
BCS
BSC
CBS
CSB
SCB
SBC

4x6x3 = 72 = 4! *3!/2! or D.
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 [#permalink] New post 20 Sep 2005, 07:43
D.. same explanation as others

For P&Cs i borrowed a book called Probability without tears from my local library and got help. i guess if u have time u can try gettin a book too
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 [#permalink] New post 20 Sep 2005, 21:22
BG wrote:
ABACUS has 6 letters, 3 vowels and 3 consonants .Take all the vowels as one unit then u have 4 elements- each of the consonants and a group of 3 vowels. This can be ordered in 4! ways. But don't forget that the group of 3 vowels can have 3! orderings within the group. Or we have 4!x3! and finally cause u have 2-A's you need to divide the whole product by2!, or your ans should be (4!x3!)/2! or option D)


Thank you very much for the simple explanation
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 [#permalink] New post 20 Sep 2005, 21:29
coffeeloverfreak wrote:
Same here, intuitively. 72 is the answer.

There are 4 places where the string of 3 vowels can appear:
VVVCCC
CVVVCC
CCVVVC
CCCVVV

The string of vowels can be arranged in 3 ways (because two of the letters are identical As):
AAU
AUA
UAA

The string of consonants can be arranged in 6 ways (because the letters are all different):
BCS
BSC
CBS
CSB
SCB
SBC

4x6x3 = 72 = 4! *3!/2! or D.


Thank you very much.
  [#permalink] 20 Sep 2005, 21:29
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