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How many combinations of 3 letters taken from letter (a,a,b,b,c,c,d) space possible?A. 12B. 13C. 35D. 36E. 56

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gmatter2012 wrote:How numerous combinations of three letters bring away from letter (a,a,b,b,c,c,d) room possible?A. 12B. 13C. 35D. 36E. 56

**Case 1: every 3 letters room different**Number of combine of 3 that have the right to be formed from 4 letter a, b, c, d = 4C3 = 4.

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**Case 2: 2 letters space the same**Number of alternatives for the letter that appears twice = 3. (a, b, or c.)Number of options for the 3rd letter = 3. (Any the the various other 3 letters.)To incorporate these options, us multiply:3*3 = 9.Total options = 4+9 = 13.The exactly answer is B.Some test-takers might find it much easier simply to compose out all of the feasible combinations:

**Case 1: every 3 letters room different**abcabdacdbcd4 options.

**Case 2: 2 letters are the same**aabaacaadbbabbcbbdccaccbccd9 options.Total options = 4+9 = 13.

Mitch HuntPrivate Tutor because that the GMAT and GRE

gmatter2012 wrote:How plenty of combinations of 3 letters bring away from letter (a,a,b,b,c,c,d) room possible?A. 12B. 13C. 35D. 36E. 56

**Case 1: all 3 letters space different**Number of combinations of 3 that have the right to be formed from 4 letters a, b, c, d = 4C3 = 4.

**Case 2: 2 letters room the same**Number of choices for the letter that appears twice = 3. (a, b, or c.)Number of alternatives for the third letter = 3. (Any that the other 3 letters.)To incorporate these options, us multiply:3*3 = 9.Total alternatives = 4+9 = 13.The exactly answer is B.Some test-takers could find it much easier simply to write out every one of the feasible combinations:

**Case 1: every 3 letters are different**abcabdacdbcd4 options.

**Case 2: 2 letters are the same**aabaacaadbbabbcbbdccaccbccd9 options.Total alternatives = 4+9 = 13.

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Thank friend Mitch ns was simply confused as to the variety of cases that have to be taken instance 1 : every are unique 4C3= 4 case 2 : 2 of one kind and also other different: 3C1*3C1= 9where 3C1 picking 1 i beg your pardon will kind the pair indigenous aa bb cc, second 3C1 selecting 1 i beg your pardon will remain after we have actually chosen the pair .lets speak we have chosen aa because that the pair then because that the solitary letter we have bb, cc, and also d so usually 3 distinct choices , because that the single letter therefore 3C1.Thank you

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I"m pretty sure that, in its existing form, this would never ever be a true GMAT question. It"s as well ambiguous. The main difficulty is the assumption that words "combination" suggests that the order of the letter does no matter. If this assumption is correct climate Mitch"s systems is perfect (as always). Having actually said that, ns don"t think this is necessarily how people interpret words "combination." for example, as soon as you use a mix lock, the bespeak of the numbers absolutely matters.As you deserve to see in Mitch"s solution, we have actually abc as one combination, however we"re not including 5 various other arrangements the the 3 letters (acb, bac, bca, cab, cba). Instead, we"re simply looking at the number of ways to choose 3 letters, there is no considering their order.To be a true GMAT question, any kind of ambiguity have to be removed. One fix:

**In how numerous different ways can 3 letter be selected from a,a,b,b,c,c,d if the order in which the letters space selected does no matter.**(I"m sure the test-maker could do far better than this)Alternatively, if the order does matter (i.e., abc, acb, bac, bca, cab and cba are thought about different), then the wording might be:

**How plenty of different 3-letter words have the right to be created if the 3 letters are selected from a,a,b,b,c,c,d?**Cheers,Brent