-
Moonlight shadow
21 photos combinations, a maths question
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?
-
Chaos
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)
-
Senior Member
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
Last edited by ctranter; 03-16-2009 at 01:03 PM.
-
Senior Member
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
Last edited by ctranter; 03-16-2009 at 01:26 PM.
-
Chaos
more info is required for this question.
-
Senior Member
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
-
Moonlight shadow
Amazing guys. I had 5,842,587,018,598,200,000,000,000,000 or 441. So yeah, nice one!
Thanks.
-
Moonlight shadow
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?
-
Moonlight shadow
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...
-
Chaos
asheep what do you mean by combinations?
-
Senior Member
I make the summation series to be:
Which my calculator says evaluates to 538,257,874,439 (5.38257874439x10^11)
-
Senior Member
My calc didn't give sufficient d.p. so I re-ran it on the pc and got exactly the same
-
He has risen!
you guys are way to fing smart to be hanging around here (except silver of course)
-
Moonlight shadow
I love the use of Flash, brilliant.
Thanks a million guys, appreciated!
Tags for this Thread
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts
-
Forum Rules
|
Click Here to Expand Forum to Full Width
|