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Re: Combinatorics - Perms [#permalink]
16 Feb 2010, 09:22

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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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: Combinatorics - Perms [#permalink]
17 Feb 2010, 02:21

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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Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: How many combinations of three letters taken from letters [#permalink]
10 Aug 2012, 16:21

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?

Re: How many combinations of three letters taken from letters [#permalink]
23 Sep 2012, 07:22

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.

Re: How many combinations of three letters taken from letters [#permalink]
25 Jan 2013, 11:36

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]
26 Jan 2013, 04:09

2

This post received KUDOS

Expert's post

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.