How many pictures would you have to take? (Much math involved)

I’m pretty sure I could figure out the answer if I sat with Microsoft Excel long enough, but I’ll save some time and bet that someone here can figure this out in their head:

If you have eight people, how many different pictures would you have to take to capture every combination of people? Meaning person A alone, person A and B together, person A and C together, all the way to all eight people in one shot. Arrangement of the people doesn’t matter; sitting A-B-C is the same as sitting B-A-C.

This came about as my mother-in-law kept taking pictures at a birthday. “OK, the three of you together, now just the two of you, now the three of you and dad.” With her arranging everyone I had plenty of time for my mind to wander. :rolleyes:

Hmm . . . that brings up a bonus question. If it takes 5 second a person per picture, how long would it take to get all those snapshots? Of course it’s more than 5 seconds if my mom-in-law has had some wine.

C(8,1) + C(8,2) + C(8,3) + … + C(8,8)

Or, 255.

This is called a Combination in mathematics.

If nobody in the picture would have been a possibility, then it would have been 2^8 = 256. A little easier to calculate.

Additionally, at 5 seconds per person per photo, the time taken would be…

(C(8,1)*1 + C(8,2)*2 + C(8,3)*3 + … + C(8,8)*8) * 5 seconds

[cc], that’s the right answer, but you got to it the hard way. The simpler way is this:

Think of every person as one bit of an eight-bit number, each bit can be either on (that person present in the picture), or off. There are 256 combinations for eight bits. Of course, you’re probably not interested in the combination where they’re all off, so that leaves 255.

An easier way to get at it is 2^8 = 256. -1 = 255 if you don’t wish to count a picture with nobody in it (C(8,0)). Each combination is determined by deciding whether each person is in or out of it. Similarly, there are 2^10 - 1 = 1023 ways bowling pins could be arranged (some rather unlikely to arise). If you like, consider each combination to match an n-bit number, disallowing 0 if you wish.

IOW, C(N,0) + C(N,1) … + C(N,N) = 2^N

… yeah, this was bound to be a dogpile, … errr … massive simulpost.

The easier way to state it (rather than using combintions) is just 2[sup]n[/sup]-1.

You could take your pictures by lining up everyone, and assigning each person a spot in that line. For any picture, you could write out who’s in it using the positions in the line - write a ‘1’ to that spot if the person is in the picture, and a ‘0’ if they aren’t.

For example, if you have four people A, B, C, D.

A B C D
1 1 1 1 = all the people are in this picture
1 0 0 0 = person A is alone in the picture
0 1 0 1 = persons B & D are in the picture.

This pattern is, in fact, a binary number. If you count in binary all numbers up to an n-length number, it’s 2[sup]n[/sup]. The -1 is because we don’t take an empty picture (corresponding to the number 0 ).

But combinations are fun!

Yep. That’s the easiest. For any problem like this, try to find a way to make it a series of “choices”. Then you can just multiply stuff together.

Here, everyone is either “present” or “not present”. So in this case, it’s 2^8. If you had 3 possible states for everyone (like "here without hat, not here, here with hat), then it’s 3^8.

More abstractly, what the OP is asking is how many different subsets are possible from a set with 8 (or more generally, n) members. The answer, as others have already pointed out, is 2[sup]n[/sup], which includes the empty set and the entire original set.

And, if you had 100 people, it would take about 200,000,000,000,000,000,000,000 years. Of course, the universe would have ended before you finished the process.

BTW, another way the OP could conceptualize it is through a recursion formula:

There is 1 way to arrange no people. If there are C[sub]n[/sub] ways to arrange n people, there are 2*C[sub]n[/sub] ways to arrange n+1 people - each of the combinations of n either with or without the (n+1)st person.

Binary! :smack:

Should I feel stupid for not having thought of that?

I could blame my father-in-law; I asked him first and, being an accountant, he started talking about making a matrix, graphing, extrapolating . . . and I tuned him out after that. Accountants give me headaches, no offense to any that happen to be here.

I’m just excited I got the (correct) answer in 4 minutes. That’s why I love this so much more than Yahoo Answers.

Something that to me is interesting is that

C(8,0) + C(8,1) + C(8,2) + C(8,3) + … + C(8,8)

is equal to 2[sup]8[/sup]. Cranking through all those combination terms, you’d never expect it.

One picture plus photoshop

For any positive integer n,

C(n,0) + C(n,1) + C(n,2) + C(n,3) + … + C(n,n) = 2[sup]n[/sup]

You can also think of it in terms of Pascal’s triangle, where each line contains the terms of such a sum.

There’s a genius in every thread.

Special case of the binomial theorem

I spent a large chunk of one year of high school math proving that, starting with combinations and permutations.