How many combinations of three letters taken from letters

Question

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

Bunuel · Accepted Answer

This kind of question has little chances appearing on the actual test.

Anyway, we have 7 letters {a, a, b, b, c, c, d}. There are 2 ways to select 3 letters out of this set:

CASE #1: all letters are distinct:

Since there are 4 distinct letters a, b, c and d, then the # of ways to select 3 out of 4 is 4C3=4.

CASE #2: 2 letters are the same and the third is different:

There are 3 letters from the set which can provide us with two letters: a, b, and c. 3C1=3 gives the # of ways to select which letter out of these 3 will provide us with 2 letters. For, example double letters can be aa, bb, or cc.

Next, we are left with 3 letters to choose the third letter. For example, if we choose aa, then b, c, and d are left to choose from for the third letter, thus the # of ways to do that is 3C1=3.

Total # of ways for this case is therefore 3C1*3C1=9.

Total for both cases = 4+9 = 13.

Answer: B.

Hope it's clear.

gixxer1000 · Answer

Im getting 13.

It says how many combinations. So each result must be different. 

Ex. you cant use abc and cba.

You can start with the the 4 seperate letters in combination of 3.

4C3=4

You can then account for multiple letters.

Ex. aab, aac, aad,bba,bbc,bbd,cca,ccb,ccd

That's another 9 combinations.

4+9=13

Ans. C

spider · Answer

I am getting 35-8 = 23 ..but none of the choices is that.

Whatever · Answer

35 - 8 will be 27 =) 

I'll bet on simplicity here 7C3 = 35

asdert · Answer

I agree on 35. Problem doesn't say letters must be different or anything else like that.

GMATBLACKBELT · Answer

7!/3!4! --> 35.

hehe 35-8=23 ---> I always make stupid errors such as this.

walker · Answer

B

4C3=4 - all different letters 
3C1*3C1=9 - two letters are the same.
N=4+9=13

hgp2k · Answer

4C3+3C1*3C1 = 13

johnnymac · Answer

jeeteshsingh - you use it to account for the following:

aa w/ b,c,d (aac, aab, aad) - 3 of these
bb w/ a,c,d -3 of these
cc w/ a,b,d -3 of these

3C1 is choose 3 letters w/ two of the same letter

It makes more sense to me as 3*3C1 = 9 total (rather than 3C1*3C1)

bangalorian2000 · Answer

7C3 = 35 hence C.

Is it really 700 plus question as mentioned in tag.

bangalorian2000 · Answer

Question does not say that we cannot select duplicate letters. So to select 3 charaters from 7 characters hence 7C3.

jeeteshsingh · Answer

This isn't correct. You need to take into account that some of the elements are duplicate. With your logic.. if u have the letters as {a,a,a,a,a,a,a}... then would no of comb possible for 3 letter word be 7c3? I don't think so.

Your answer would only be correct if all the 7 letters were different. If anyone of them repeated, the combinations would become less in number. Please check!

magmag · Answer

I always learn by modifying problems a little bit and see what answers I can come up with to test my understanding so...

If I assume that there's an extra a, an extra b, and an extra c for example. i.e. We have (a,a,a,b,b,b,c,c,c,d). Will that change our computation of 4C3+3C1*3C1? I myself don't think so but I'm waiting for your comments.

If I assume that's there's an an extra couple of D's and E's and 1 F i.e. (a,a,b,b,c,c,d,d,d,e,e,f). In this case the answer, as I guess, is 6C3+5C3*4C1=60. Am I correct?

Thanks

anilisanil · Answer

Just got confused.

approaching by filling in slots.

we have three slots for seven letters.

now the first slot can be filled by any of the 7 letters, then 6, then 5.

so if there were 7 different letters, we would have had 7*6*5 and remove the repetitive combinations by dividing with 3! that makes 35 combinations.

For this approach, I got stuck here not knowing how to delete the repetitive combinations due to double letters. Help please.

P.S: I got the answer through other approach by adding unique and double letter combinations, I just want to understand why I got stuck above, rather how to continue from above.

manimgoindown · Answer

Hey could someone please explain how they got 3C1 3C1? There are only two elements chosen in that situation? There doesn't seem to be a good explanation, or an OA for this problem

sambam · Answer

Combinations only... total 13 (answer B)
So 4C3 for abcd - 4 choices
3 more for aa(b,c or d)
3 more for bb(a,c or d)
3 more for cc(a,b or d)

total 13

email2vm · Answer

Sol:

consider any one pair with d 
one pair from 3 pairs = 3c1 *1 (AAD or BBD or CCD)
But they can interchange their position(AAD,ADA,DAA) hence 3c1*1*3!/2! = 9

also we can consider all different digits

so we have 4 distinct objects and we have to choose 3 i.e. 4c3 = 4

9+4=13

Vinitkhicha1111 · Answer

Indeed writing it as : 2C2*3C1*3=9 will make the most sense and clear it to almost anybody. ( 2C2 refers to the choosing of same 2 letters and then 3C1 refers to chossing any one letter from among the 3 diff types of letters left and then finally *3 gives the total such possible discrete cases.) Hope this helps!!

Anjalika123 · Answer

Hi Bunnel ,
Let -->s (same letter)
 --> d(distinct letter)
Then
In case 2(two same other distinct) shouldn't we consider these cases?
-->ssd >>3*3=9
-->sds >>3*3=9
--> dss >>3*3=9
so # of possibilities for case two =9*3=27

How many combinations of three letters taken from letters

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How many combinations of three letters taken from letters

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