How many combinations of three letters taken from letters

[gmatter2012]

How many combinations of three letters taken from letters (a,a,b,b,c,c,d) are possible?

A. 12
B. 13
C. 35
D. 36
E. 56

[GMATGuruNY]

How many combinations of three letters taken from letters (a,a,b,b,c,c,d) are possible?

A. 12
B. 13
C. 35
D. 36
E. 56

Case 1: All 3 letters are different
Number of combinations of 3 that can be formed from 4 letters {a, b, c, d} = 4C3 = 4.

Case 2: 2 letters are the same
Number of options for the letter that appears twice = 3. (a, b, or c.)
Number of options for the third letter = 3. (Any of the other 3 letters.)
To combine these options, we multiply:
3*3 = 9.

Total options = 4+9 = 13.

The correct answer is B.

Some test-takers might find it easier simply to write out all of the possible combinations:

Case 1: All 3 letters are different
abc
abd
acd
bcd
4 options.

Case 2: 2 letters are the same
aab
aac
aad
bba
bbc
bbd
cca
ccb
ccd
9 options.

Total options = 4+9 = 13.

[gmatter2012]

How many combinations of three letters taken from letters (a,a,b,b,c,c,d) are possible?

A. 12
B. 13
C. 35
D. 36
E. 56

Case 1: All 3 letters are different
Number of combinations of 3 that can be formed from 4 letters {a, b, c, d} = 4C3 = 4.

Case 2: 2 letters are the same
Number of options for the letter that appears twice = 3. (a, b, or c.)
Number of options for the third letter = 3. (Any of the other 3 letters.)
To combine these options, we multiply:
3*3 = 9.

Total options = 4+9 = 13.

The correct answer is B.

Some test-takers might find it easier simply to write out all of the possible combinations:

Case 1: All 3 letters are different
abc
abd
acd
bcd
4 options.

Case 2: 2 letters are the same
aab
aac
aad
bba
bbc
bbd
cca
ccb
ccd
9 options.

Total options = 4+9 = 13.

Thank you Mitch

I was just confused as to the number of cases that need to be taken

case 1 : all are distinct 4C3= 4
case 2 : two of one kind and other different: 3C1*3C1= 9

where 3C1 choosing 1 which will form the pair from aa bb cc,
second 3C1 choosing 1 which will remain after we have chosen the pair .

lets say we have chosen aa for the pair then for the single letter we have bb, cc, and d so basically 3 distinct choices , for the single letter hence 3C1.

Thank you

[eski]

Is this a right GMAT question and of what difficulty?

[Brent@GMATPrepNow]

Is this a right GMAT question and of what difficulty?

I'm pretty sure that, in its current form, this would never be a true GMAT question. It's too ambiguous.
The main problem is the assumption that the word "combination" implies that the order of the letters does not matter. If this assumption is correct then Mitch's solution is perfect (as always).

Having said that, I don't think this is necessarily how people interpret the word "combination." For example, when you use a combination lock, the order of the numbers certainly matters.

As you can see in Mitch's solution, we have abc as one combination, but we're not including 5 other arrangements of the 3 letters (acb, bac, bca, cab, cba). Instead, we're just looking at the number of ways to select 3 letters, without considering their order.

To be a true GMAT question, any ambiguity must be removed.

One fix: In how many different ways can 3 letters be selected from {a,a,b,b,c,c,d} if the order in which the letters are selected does not matter. (I'm sure the test-maker could do better than this)

Alternatively, if the order does matter (i.e., abc, acb, bac, bca, cab and cba are considered different), then the wording could be: How many different 3-letter words can be constructed if the 3 letters are selected from {a,a,b,b,c,c,d}?

Cheers,
Brent