Using the standard 26-letter alphabet, how many different groups of 4 letters are possible, if exactly two letters are chosen from a, e, i, o ,u and two letters are chosen from all the remaing letters except for y, and a letter can be used more than once?


 (A) 500
 (B) 1900
 (C) 2500
CORRECT  (D) 3150
 (E) 6300
Recognize the GMAT Topic Being Tested
This is a combination, in which order doesn’t matter. 

Write Down the GMAT Information NOT Given in the Question

First, we draw and label the number of spots. Ultimately we will multiply across the tops and divide by the subscripts at the bottom. Let’s use v for the vowels and c for the consonants.





Connect the Information Not Given to the Information Given

Normally, the next step would be to fill in the number of choices for each spot. However, this question deviates from the usual combination because the vowels and consonants can each appear twice. One way to handle this attribute is to solve without replacement and then add in the number of additional combinations that occur by doubling each vowel and/or consonant.

We can think of the above equation as having two separate components, one for the vowels and one for the consonants.

/mstart({{}\over {v1}}\bullet{{}\over {v2}})\bullet({{}\over {c1}}\bullet{{}\over {c2}})/mend

Let’s break them apart and deal with each individually. We will have to solve each as a normal combination without doubling any options and then add in the doubles.


Solve

For the vowels,

/mstart{{}\over {v1}}\bullet{{}\over {v2}}/mend

Without reusing vowels, there would be five choices for the first spot and four for the second spot. Since order doesn’t matter, we must divide by the value of each subscript.

/mstart({{5}\over {v1}}\bullet{{4}\over {v2}})=({{5}\over {1}}\bullet{{4}\over {2}})=10/mend

There are ten possible combinations where no vowel is repeated. But with repeats, there are five additional choices: aa, ee, ii, oo and uu.

Thus, the total number of vowel combinations is 10 + 5 = 15

For the consonants,

/mstart{{}\over {c1}}\bullet{{}\over {c2}}/mend

Without reusing consonants, there would be twenty choices for the first spot and nineteen for the second spot(Don’t forget not to count y). Again here, since order doesn’t matter, we must divide by the value of each subscript.

/mstart({{20}\over {c1}}\bullet{{19}\over {c2}})=({{20}\over {1}}\bullet{{19}\over {2}})=190/mend


There are 190 possible combinations where no vowel is repeated. But with repeats, there are 20 additional choices.

Thus, the total number of vowel combinations is

190 + 20 = 210.

Now we can multiply 15 by 210 to get the total number of combinations, which is 3150.


Reread the Question Asked and Select an Answer
The question asked us for how many groups of letters we can choose and we solved for that.

Choose D.
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