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counting [#permalink] New post 27 Nov 2010, 06:11
00:00
A
B
C
D
E

Difficulty:

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Question Stats:

56% (01:45) correct 44% (02:20) wrong based on 9 sessions
How many different groups of 4 letters are possible, if exactly two vowels (not including y as a vowel) and two consonants must be among the letters selected, a letter can be used more than once, and all letters are available but the letter Q?


(A) 125
(B) 250
(C) 500
(D) 2500
(E) 15000
[Reveal] Spoiler: OA

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Re: counting [#permalink] New post 28 Nov 2010, 02:00
Choosing 2 vowels out of five with repetitions = 5*5
choosing 2 consonants out of 20 (26-5 vowels-Q) =20*20

and since we need different arrangements we need to divide by 2!*2!

=5*5*20*20/2!*2!= 2500
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Re: counting [#permalink] New post 03 Dec 2010, 02:57
can someone give another approach
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Re: counting [#permalink] New post 22 Feb 2011, 10:48
can someone pls explain why division by 2!*2!..................? pls explain with example if possible ....not able to get it ......
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Re: counting [#permalink] New post 22 Feb 2011, 11:51
AkritiMehta wrote:
Choosing 2 vowels out of five with repetitions = 5*5
choosing 2 consonants out of 20 (26-5 vowels-Q) =20*20

and since we need different arrangements we need to divide by 2!*2!

=5*5*20*20/2!*2!= 2500


You might look at your calculation just for the selection of the vowels alone; using this method, there would be 5*5/2! = 12.5 ways to pick the two vowels. The answer needs to be a whole number, so that can't be right! I'm guessing, since the OA is quoted as 2500, that the source used the same method as you did, but it isn't correct. The right answer to the question in the original post isn't among the answer choices, so I wonder where the question is from - I wouldn't use the source for anything else.

It might be easiest to see how to count here by starting with the simplest case. Say we have two letters A, B, and we want to know how many groups of 2 letters we can pick if repetition is allowed and if order is not important. We can list the 3 possibilities:

A, A
B, B
A, B

When the letters are different, we need to divide by 2, because we don't want to count {A, B} and {B, A} twice if order does not matter. But when the letters are identical, we do not want to divide by 2, since we aren't double-counting anything. So for two letters, we have 2 ways to choose the set of letters when the letters are the same, and 2C2 = 1 way of choosing the letters if they are different.

Going back to the original question and proceeding similarly, if we choose 2 vowels from the group of 5, we have 5 ways of choosing two identical vowels, and 5C2 = 10 ways of choosing 2 different vowels, for a total of 5 + 10 = 15 ways to choose two vowels. For the 20 consonants, we have 20 ways of choosing two identical consonants, and 20C2 = 190 ways of choosing 2 different consonants, for a total of 20+190 = 210 ways to choose two consonants. Multiplying the choices for vowels and consonants gives a total of 15*190 = 3150 ways of choosing a group of four letters with the restrictions given.
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Re: counting [#permalink] New post 23 Feb 2011, 10:34
hmm, lets say i want to do it like this:
5C2*20C2

i need to divide it by how much?

thanks.
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Re: counting   [#permalink] 23 Feb 2011, 10:34
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