21 photos combinations, a maths question

[asheep_uk]

I'm a little stuck on the maths of this puzzle. Anybody know?

I have 21 photos and I've given them to a few dozen different people. Each person can order them how they like and use as many or as little as they wish, but may not use the same one twice.

How many different combinations are there?

[silverx2]

y number of people
x number of combinations per person
u number of combos total
x * y = u

some info that is needed is what do you mean may not use the same one twice.

if you send one person 21 pictures, he has at the most 21 different combinations if he can only use a single picture one time.

so that would make it

21*Y=U

but if you are talking combinations of photos such as, they cant use pictures 1 and 3 together more then once, then it turns into

21+20+19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4+3+ 2+1=x (i dont know how to make the correct mathmatical symbol in a forum)

[ctranter]

You can think of it as a summation series of combinations.

nCr is the number of ways of choosing r things from a total of n
so here n=21

if people can choose as many photos they like up to a total of 21, and as few as 1 photo
then we have the series

21C21 + 21C20 + 21C19 ... 21C2 + 21C1 + 21C0

or

nCn + nC(n-1) + nC(n-2) ...

A series like this converges to equal 2^n

Now because we dont want to count the option where no photos are taken the answer is (2^n) - n

or in this case = 2,097,131

Thats my take, its been years since I've done any prob theory so I might be wrong

[ctranter]

our answers are one different. annoying! i hope its close enough for asheep

actually I think it should be (2^n) - 1

as 21C0 is of course 1 not 21

and hence 2,097,151

hmmmmm

[silverx2]

more info is required for this question.

[ctranter]

yeah your adding is at fault jtnw

21
210
1330
5985
20349
54264
116280
203490
293930
352716
352716
293930
203490
116280
54264
20349
5985
1330
210
21
1 +
--------
2097151

[asheep_uk]

Amazing guys. I had 5,842,587,018,598,200,000,000,000,000 or 441. So yeah, nice one!

Thanks.

[asheep_uk]

Crap, now I need to work out how many combinations there are if you're allowed to use the same photo more than once (so I can compare...)

That must be quite simple?

[asheep_uk]

If I do that I end up with 5,842,587,018,598,200,000,000,000,000 (or 5.84258701838598e+27) Formula being n^r. That seems like a big number though...

[silverx2]

asheep what do you mean by combinations?

[ctranter]

I make the summation series to be:



Which my calculator says evaluates to 538,257,874,439 (5.38257874439x10^11)

[ctranter]

My calc didn't give sufficient d.p. so I re-ran it on the pc and got exactly the same

[lefteyewilly]

you guys are way to fing smart to be hanging around here (except silver of course)

[asheep_uk]

I love the use of Flash, brilliant.

Thanks a million guys, appreciated!