# Math experts, help me solve this problem (CC-Combinations content)

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I have a group of 18 girls. I want to know how many possible combinations of those girls I could make for photos. (It's a theoretical question, I don't plan to do it :laugh: )

So I would have one photo of all 18 girls. Then I would have 17 groups (I think), each group with one girl missing. Then I would have a bunch of groups of 16 girls, on down to having a photo of each individual girl.

Is it 18! ?

Thanks!

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I have a group of 18 girls. I want to know how many possible combinations of those girls I could make for photos.

So I would have one photo of all 18 girls. Then I would have 17 groups (I think), each group with one girl missing. Then I would have a bunch of groups of 16 girls, on down to having a photo of each individual girl.

Is it 18! ?

Please specify: does it matter in which order the girls are in the photo, or is girls of the same photo group switching spot considered ONE choice?

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I have a group of 18 girls. I want to know how many possible combinations of those girls I could make for photos. (It's a theoretical question, I don't plan to do it :laugh: )

So I would have one photo of all 18 girls. Then I would have 17 groups (I think), each group with one girl missing. Then I would have a bunch of groups of 16 girls, on down to having a photo of each individual girl.

Is it 18! ?

Thanks!

No, it's not 18!. You could add up all the 17 groups, 16 groups, and so forth, but there's actually an easier way to think about this one. For each girl, ask yourself, is she in the picture or out of the picture. Two choices for each girl. Doing that all possible ways for each girl will give you every possible picture combination. Two independent choices for each of 18 girls is 2^18. Now that includes the picture where every girl is OUT of the picture. Probably you wouldn't include that choice, so 2^18-1 = 262,143.

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Please specify: does it matter in which order the girls are in the photo, or is girls of the same photo group switching spot considered ONE choice?

No, order doesn't matter.

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No, it's not 18!. You could add up all the 17 groups, 16 groups, and so forth, but there's actually an easier way to think about this one. For each girl, ask yourself, is she in the picture or out of the picture. Two choices for each girl. Doing that all possible ways for each girl will give you every possible picture combination. Two independent choices for each of 18 girls is 2^18. Now that includes the picture where every girl is OUT of the picture. Probably you wouldn't include that choice, so 2^18-1 = 262,143.

I had considered this, but then I tried writing it out with just 5 girls: A,B,C,D, and E

I got

ABCDE

ABCD ABCE ABDE ACDE BCDE

ABC ABD ABE BCD BCE CDE

AB AC AD AE BC BD BE CD CE DE

A B C D E

For a total of 27 groups. But using binomial coefficients that would be 2^4 = 16, which isn't right. What am I missing?

Obviously, 5! didn't work either. :o

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I had considered this, but then I tried writing it out with just 5 girls: A,B,C,D, and E

I got

ABCDE

ABCD ABCE ABDE ACDE BCDE

ABC ABD ABE BCD BCE CDE

AB AC AD AE BC BD BE CD CE DE

A B C D E

For a total of 27 groups. But using binomial coefficients that would be 2^4 = 16, which isn't right. What am I missing?

It would be 2^5 = 32. You're just missing a few of the 3-combinations. One thing you might notice is that the 3-cominations and the 2-combinations should be the same in number. They will be complements of each other.

So that brings you to 31. And then the last one is the one with NO girls in it, the 0-combination. You probably wouldn't make a picture like that, so just 2^5-1 = 31.

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It would be 2^5 = 32. You're just missing a few of the 3-combinations. One thing you might notice is that the 3-cominations and the 2-combinations should be the same in number. They will be complements of each other.

So that brings you to 31. And then the last one is the one with NO girls in it, the 0-combination. You probably wouldn't make a picture like that, so just 2^5-1 = 31.

Thank you, thank you, thank you! I knew I must have been leaving something out.

Duh. Loved stats. Hated probablility.

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