I feel like I should know this but I can't figure it out and it's...

[Eihli]

making me feel stupid.

I have 7 different items. I put them in groups of 3. Each of the 7 items is repeatable. The order of the 3 items in the group doesn't matter (AAB is the same thing as ABA and BAA, they count as 1 group, not 3).

What is the equation to figure out how many different groups I can have?
And believe it or not this isn't a problem for homework. It all started as a dispute over a game of Warcraft III.

[MarkD]

7 choose 3.
n!/(k!*(n-k)!) where n=7 and k=3 and where ! represents a factorial.

ie. 10! = 10*9*8*7*6*5*4*3*2*1 and in more general...
N! = N*(N-1)*(N-2)*...*3*2*1

[MarkD]

Wait... are you essentially asking this:

I have 7 different balls in a bag, I want to select three, after each selection I replace the ball which allows me to choose the ball again. How many different groups of 3 can I choose?

If you are asking that then the answer is different than what I said before and it's a high school question but damn if I can remember how to do it. I'll think about it more if no one else answers.

[MarkD]

If I have n balls in a bag and sample with replacement then if I choose k balls I can have n^k possible combinations.

So... 7^3 = 7*49 = 343.

[daryn]

nevermind!

[bigpooch]

For all the items the same, there are 7 possibilities.

For two of one and one of another (e.g., AAB), there are
7x6=42 possibilities.

For three distinct items there are C(7,3)=7x6x5/6=35
possibilities.

Altogether, there are 7+42+35=84 different kinds of
groupings.

I wouldn't call these sets or groups as these two terms have
a clear precise mathematical definition!

[Eihli]

111
112
113
114
115
116
117

221
222
223
224
225
226
227

331
332
333
334
335
336
337

441
442
443
444
445
446
447

551
552
553
554
555
556
557

661
662
663
664
665
666
667

771
772
773
774
775
776
777



That's 49. Did I miss some?

[bigpooch]

You are missing the 35 that have three different digits!

[Eihli]

Ever spend a few hours trying to fix your computer because it won't boot up and then you call tech support and they tell you to take the floppy disk out of the drive? That's how I feel right now /images/graemlins/frown.gif

[RocketManJames]

Maybe I'm not reading the question right, and if not ignore this.

I think this sounds like a "multi-choose" problem.

N multichoose K = (N+K-1) choose K

So in your problem you have 7 multichoose 3, which is (9 choose 3) = 84

You can prove the multichoose equation thinking of ways to permute |'s and O's.

In your case you have A | B | C | D | E | F | G

Now, think of it as having 6 |'s and 3 O's.

So, possible groupings you could have are:

| OO | | | O | |
O | O | | | | | O
OOO | | | | | |

And by showing the ways you can permute |'s and O's you can come up with the multichoose formula (N+K-1) choose K.

-RMJ

[bigpooch]