Help with Combinations

  • Thread starter Omid
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  • #1
182
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How many combinations of three letters from the letters A,A,B,B,C,C,D are ther?
------------------------------
I assumed 2 cases:
1. 2 letters are the same.
2. all the 3 are different.

And found 9 forms for the first case, and 24 for the second one. As you know my answer (33) is 20 more than what is written in the book.
 

Answers and Replies

  • #2
1,036
1
You are right about the case where 2 letters are the same.

But if you assume that all 3 are different, it is the same as selecting 3 items from a set of 4 with no repetition. This is just C(4,3) = 4!/3! = 4. Specifically, in this case, {ABC, ABD, ACD, BCD}. What other possibilities are there?

How did you get 24?
 
  • #3
182
0
I'm a bit confused. I simply multiplied in the first case:
(3 ways to choose the first 2 letters that are the same) * (3 ways to choose the third letter) = 9
And did the same in the second case:
(4 ways to choose the first letter) * (3 ways.... the second) * (2 ways... the third)= 24

I did the same thing in the both cases, but only the first one is right. Can you explain it to me?
Thanks
 
  • #4
1,036
1
When you did this
And did the same in the second case:
(4 ways to choose the first letter) * (3 ways.... the second) * (2 ways... the third)= 24
you were counting permutations, not combinations. For example, you counted each of ABC, ACB, BAC, BCA, CAB, CBA, when really they all represent the same combination of letters. So you have to divide your 24 permutations by the number of ways that 3 items can be ordered, which is 3!, or 6. 24/6=4 is the number of combinations.

The first case was different. There you were selecting a combination of a pair and a single letter. There were 3 possible pairs, and for each pair, 3 possibilities for the single letter. So when you multiplied 3 x 3 you counted each combination only once.
 
  • #5
182
0
Thank you very much.
 

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