Calculate Possible Combinations of Five Letters

I need to calculate all the possible combinations of the letters A, B, C, D, E.

I think there is a formula for this, but I can recall it.

In other words...  My output would be something like this:

A B C D E
A C D E
A D E
A E
A
A B D E
A B E
A C
A D
A E
B C D E
B D E
B E
B
C D E
C E
C

Hopefully you get the idea.

Thanks in advance!
Senniger1Asked:
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sdstuberCommented:
The links above only show for a particular length/sample pair.

You would need to repeat it for each length 1-N where N is the length of the string.

C(5,5) + C(5,4) + C(5,3) + C(5,2) + C(5,1) = 1 + 5 + 10 + 10 + 5 = 31
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Senniger1Author Commented:
Thank you for the response, but I saw all that stuff when I did a google search.  Of those, I didn't see any that showed the output where sometimes there's a 5 letter combination, sometimes a 4 letter combination, sometimes there's a 3 letter combination, etc.

I need someone to show me the calculations with my letters and show me the answer.

Many thanks!
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Senniger1Author Commented:
That was what I was looking for.  Thank you so much!
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aadihCommented:
He asked for a formula. I gave a website that has the formula. No blind link. Sorry (to disagree).
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ozoCommented:
2^5-1
(assuming you're not counting the combination with no letters)
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aadihCommented:
Again, I disagree -- respectfully -- in this particular instance.

Thanks. Regards.
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d-glitchCommented:
You are essentially asking for all the subsets of [A B C D E]

One way to think about the problem is count in binary from

00000  = 0  ==> None/Empty Set
00001  = 1  ==> E     If there is a 1 in a position, include the letter
00010  = 2  ==> D
00011  = 3  ==> DE
    :
    :
11110  = 30  ==> ABCD
11111  = 31  ==> ABCDE

This is the same result as ozo's formula.
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aadihCommented:
Nice explanation, d-glitch. Thanks.
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sdstuberCommented:
Senniger1,

I concur, with ozo's and d-glitch's posts.

While I showed how the individual combination counts sum to produce the number you're looking for, and do produce the correct answer,  I left out the next obvious step of showing the simplification.

Using the exponentiation is the much more efficient means of getting the result.

As a TA I can reopen the question so you can split points to her if you'd like.
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aadihCommented:
I believe points are not that important -- at least for me. (Individuals may differ.) But the satisfaction (that comes) in providing help or trying to help.  :-)
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Senniger1Author Commented:
If you want to reopen the questions I will split some of the points as I deem fair since you answered my question as to what I needed.
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aadihCommented:
Nice Charitable.  :-)
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Senniger1Author Commented:
I based the points on the first response which completely answered my question and the detail which was most useful to me.

Many thanks to all!
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