Author 
Message 
TAGS:

Hide Tags

CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

How many combinations of three letters taken from letters [#permalink]
Show Tags
02 Dec 2007, 14:29
5
This post received KUDOS
27
This post was BOOKMARKED
Question Stats:
30% (02:40) correct
70% (01:39) wrong based on 721 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Senior Manager
Joined: 06 Aug 2007
Posts: 361

Re: Combinatorics  Perms [#permalink]
Show Tags
02 Dec 2007, 18:58
bmwhype2 wrote: 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
I am getting 358 = 23 ..but none of the choices is that.



Manager
Joined: 03 Sep 2006
Posts: 233

Re: Combinatorics  Perms [#permalink]
Show Tags
03 Dec 2007, 01:35
spider wrote: bmwhype2 wrote: 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 I am getting 358 = 23 ..but none of the choices is that.
35  8 will be 27 =)
I'll bet on simplicity here 7C3 = 35



Senior Manager
Joined: 09 Oct 2007
Posts: 466

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



CEO
Joined: 29 Mar 2007
Posts: 2559

Re: Combinatorics  Perms [#permalink]
Show Tags
03 Dec 2007, 22:36
1
This post received KUDOS
bmwhype2 wrote: 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
7!/3!4! > 35.
hehe 358=23 > I always make stupid errors such as this.



Director
Joined: 26 Jul 2007
Posts: 535
Schools: Stern, McCombs, Marshall, Wharton

8
This post received KUDOS
1
This post was BOOKMARKED
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



CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: Combinatorics  Perms [#permalink]
Show Tags
03 Jan 2008, 16:44
4
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
B4C3=4  all different letters 3C1*3C1=9  two letters are the same. N=4+9=13
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



Senior Manager
Joined: 18 Aug 2009
Posts: 320

Re: Combinatorics  Perms [#permalink]
Show Tags
27 Sep 2009, 22:15
2
This post received KUDOS
\(4C3+3C1*3C1 = 13\)



Manager
Joined: 29 Nov 2009
Posts: 107
Location: United States

Re: Combinatorics  Perms [#permalink]
Show Tags
16 Feb 2010, 10:49
2
This post received KUDOS
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)



Senior Manager
Joined: 01 Feb 2010
Posts: 264

Re: Combinatorics  Perms [#permalink]
Show Tags
16 Feb 2010, 11:09
bmwhype2 wrote: 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 7C3 = 35 hence C. Is it really 700 plus question as mentioned in tag.



Senior Manager
Joined: 01 Feb 2010
Posts: 264

Re: Combinatorics  Perms [#permalink]
Show Tags
16 Feb 2010, 22:59
johnnymac wrote: The answer is 13, not 35. Question does not say that we cannot select duplicate letters. So to select 3 charaters from 7 characters hence 7C3.



Senior Manager
Joined: 22 Dec 2009
Posts: 359

Re: Combinatorics  Perms [#permalink]
Show Tags
17 Feb 2010, 03:21
bangalorian2000 wrote: johnnymac wrote: The answer is 13, not 35. Question does not say that we cannot select duplicate letters. So to select 3 charaters from 7 characters hence 7C3. 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!
_________________
Cheers! JT........... If u like my post..... payback in Kudos!!
Do not post questions with OAPlease underline your SC questions while postingTry posting the explanation along with your answer choice For CR refer Powerscore CR BibleFor SC refer Manhattan SC Guide
~~Better Burn Out... Than Fade Away~~



Intern
Joined: 20 Oct 2011
Posts: 12

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
10 Aug 2012, 17:21
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



Manager
Joined: 18 Aug 2006
Posts: 113
Location: United States
WE: Consulting (Telecommunications)

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
23 Sep 2012, 08:22
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.



Manager
Joined: 07 Feb 2011
Posts: 105

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
25 Jan 2013, 12:36
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
_________________
We appreciate your kudos'



Manager
Joined: 18 Oct 2011
Posts: 90
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01302013
GPA: 3.3

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
25 Jan 2013, 14:11
1
This post received KUDOS
1
This post was BOOKMARKED
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



Math Expert
Joined: 02 Sep 2009
Posts: 39750

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
26 Jan 2013, 05:09
11
This post received KUDOS
Expert's post
9
This post was BOOKMARKED
bmwhype2 wrote: 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 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16022

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
26 Jul 2014, 22:54
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Manager
Status: folding sleeves up
Joined: 26 Apr 2013
Posts: 158
Location: India
Concentration: Finance, Strategy
GMAT 1: 530 Q39 V23 GMAT 2: 560 Q42 V26
GPA: 3.5
WE: Consulting (Computer Hardware)

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
24 Sep 2014, 07:57
bmwhype2 wrote: 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 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



Manager
Joined: 18 Jun 2014
Posts: 82
Concentration: General Management, Finance
Schools: Stanford '16, Wharton '16, Booth PT '16, Sloan '16, CBS '16, Haas '16, Tuck '16, Stern '16, Duke '16, LBS '16, Insead Sept '16, ISB '16
GMAT 1: 720 Q50 V38 GMAT 2: 740 Q50 V40
GPA: 3.8
WE: Management Consulting (Insurance)

Re: How many combinations of three letters taken from letters [#permalink]
Show Tags
06 Oct 2014, 05:44
johnnymac wrote: 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) 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!!
_________________
The Mind is everything . What you think you become.  Lord Buddha
Consider giving KUDOS if you appreciate my post !!




Re: How many combinations of three letters taken from letters
[#permalink]
06 Oct 2014, 05:44



Go to page
1 2
Next
[ 24 posts ]




