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How many combinations of three letters taken from letters [#permalink]
 02 Dec 2007, 14:29
Question Stats:
20% (01:58) correct
79% (01:17) wrong based on 94 sessions
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
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Re: Combinatorics - Perms [#permalink]
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 35-8 = 23 ..but none of the choices is that.
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Re: Combinatorics - Perms [#permalink]
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 35-8 = 23 ..but none of the choices is that. 35 - 8 will be 27 =)
I'll bet on simplicity here 7C3 = 35
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I agree on 35. Problem doesn't say letters must be different or anything else like that.
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Re: Combinatorics - Perms [#permalink]
03 Dec 2007, 22:36
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 35-8=23 ---> I always make stupid errors such as this.
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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
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Re: Combinatorics - Perms [#permalink]
03 Jan 2008, 16:44
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B4C3=4 - all different letters 3C1*3C1=9 - two letters are the same. N=4+9=13
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Re: Combinatorics - Perms [#permalink]
03 Jan 2008, 22:52
walker wrote: B
4C3=4 - all different letters
3C1*3C1=9 - two letters are the same.
N=4+9=13 Thx man. Kinda figured that you can't run combination formula on a multiset. Is there a generic formula for combinations of a multiset?
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Re: Combinatorics - Perms [#permalink]
27 Sep 2009, 22:15
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4C3+3C1*3C1 = 13
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Re: Combinatorics - Perms [#permalink]
16 Feb 2010, 10:22
walker wrote: B
4C3=4 - all different letters
3C1*3C1=9 - two letters are the same.
N=4+9=13 Hi Walker.... care to explain the highlighted part in detail..... guess ma brain has switched off.
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Re: Combinatorics - Perms [#permalink]
16 Feb 2010, 10:49
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)
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Re: Combinatorics - Perms [#permalink]
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.
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Re: Combinatorics - Perms [#permalink]
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.
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Re: Combinatorics - Perms [#permalink]
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!
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Re: How many combinations of three letters taken from letters [#permalink]
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
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Re: How many combinations of three letters taken from letters [#permalink]
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.
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Re: How many combinations of three letters taken from letters [#permalink]
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
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Re: How many combinations of three letters taken from letters [#permalink]
25 Jan 2013, 14:11
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
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Re: How many combinations of three letters taken from letters [#permalink]
26 Jan 2013, 05:09
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 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.
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Re: How many combinations of three letters taken from letters
[#permalink]
26 Jan 2013, 05:09
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