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In how many ways can a 4-letter word be formed from the

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In how many ways can a 4-letter word be formed from the

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In how many ways can a 4-letter word be formed from the letters ABCDEFIO such that the word contains 2 vowels and 2 consonants? The letters cannot be repeated, and, the words can have no dictionary meaning.

A. 36
B. 144
C. 288
D. 864
E. 1728

The OA is D

Source: e-GMAT

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swerve wrote:
In how many ways can a 4-letter word be formed from the letters ABCDEFIO such that the word contains 2 vowels and 2 consonants? The letters cannot be repeated, and, the words can have no dictionary meaning.

A. 36
B. 144
C. 288
D. 864
E. 1728
Source: e-GMAT
$$?\,\,:\,\,\# \,\,4\,\,{\rm{distinct}}\,\,{\rm{letters}}\,\,\,\left\{ \matrix{
\,2\,\,{\rm{vowels}} \hfill \cr
\,2\,\,{\rm{consonants}} \hfill \cr} \right.$$
$$?\,\,\, = \,\,\,\underbrace {C\left( {4,2} \right)}_{{\rm{for}}\,\,{\rm{vowels}}} \cdot \underbrace {C\left( {4,2} \right)}_{{\rm{for}}\,\,{\rm{consonants}}} \cdot \underbrace {\,\,{P_4}\,\,}_{{\rm{chosen}}\,4\,,\,\,{\rm{permutations}}}\,\,\, = \,\,\,{{4 \cdot 3} \over 2} \cdot {{4 \cdot 3} \over 2} \cdot 4!\,\,\, = \,\,\,6 \cdot 6 \cdot 24\,\,\, = \,\,\,864$$

The correct answer is therefore (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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swerve wrote:
In how many ways can a 4-letter word be formed from the letters ABCDEFIO such that the word contains 2 vowels and 2 consonants? The letters cannot be repeated, and, the words can have no dictionary meaning.

A. 36
B. 144
C. 288
D. 864
E. 1728
There are 4C2 = 6 ways to choose 2 vowels from 4 vowels, and similarly, there are 4C2 = 6 ways to choose 2 consonants from 4 consonants Therefore, there are 6 x 6 = 36 different words with a particular set of 2 vowels and 2 consonants.

However, for each of these 36 words, since the 4 letters can be arranged in any order, there are a total of 4! x 36 = 24 x 36 = 864 6 words possible.

Answer: D

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swerve wrote:
In how many ways can a 4-letter word be formed from the letters ABCDEFIO such that the word contains 2 vowels and 2 consonants? The letters cannot be repeated, and, the words can have no dictionary meaning.

A. 36
B. 144
C. 288
D. 864
E. 1728

The OA is D

Source: e-GMAT
BIDE....BODE...etc $$$$

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Hi regor60,

I agree that this prompt is poorly-worded. The "intent" is to ask for every 4-letter arrangement that fits the given restrictions, INCLUDING "words" that do not appear in the dictionary (re: arrangements that are not actually words). GMAT question-writers are far more specific with how they word their questions, so you won't face this type of ambiguity on Test Day.

GMAT assassins aren't born, they're made,
Rich

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Rich.C@EMPOWERgmat.com wrote:
Hi regor60,

I agree that this prompt is poorly-worded. The "intent" is to ask for every 4-letter arrangement that fits the given restrictions, INCLUDING "words" that do not appear in the dictionary (re: arrangements that are not actually words). GMAT question-writers are far more specific with how they word their questions, so you won't face this type of ambiguity on Test Day.

GMAT assassins aren't born, they're made,
Rich
Gotcha.

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swerve wrote:
In how many ways can a 4-letter word be formed from the letters ABCDEFIO such that the word contains 2 vowels and 2 consonants? The letters cannot be repeated, and, the words can have no dictionary meaning.

A. 36
B. 144
C. 288
D. 864
E. 1728
Alternate approach:

Case 1: VVCC, where the first two letters are vowels and the last 2 letters are consonants
Number of options for the first vowel = 4. (Any of the 4 vowels)
Number of options for the second vowel = 3. (Any of the 3 remaining vowels)
Number of options for the first consonant = 4. (Any of the 4 consonants)
Number of options for the second consonant = 3. (Any of the 3 remaining consonants)
To combine these options, we multiply:
4*3*4*3 = 144.

Other cases:
To account for every possible position for the two selected vowels and the two selected consonants, we must multiply the result above by the number of ways to arrange VVCC.
The number of ways to arrange 4 distinct letters = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be arranged.
The reason:
When identical elements swap positions, the arrangement does not change.
Here, we must divide by 2! to account for VV and by another 2! to account for CC:
4!/(2!2!) = 6.

Thus:
Total number of possible words = 144 * 6 = 864.

The correct answer is D.

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