Curse

Yellow_lid · Oct 07, 2007, 07:34 PM // 19:34

Generic disclaimer. (See Q1)

I spent 8 hours solid on this problem yesterday and I finally solved it WITHOUT the use of internet help. Start from scratch and solve the problem. There is a more general solution and that will be correct, IF you solve the original problem with your general solution.

The Question is:
In how many ways can you distribute nine different candies to three children in such a way that each child gets at least one candy?

If nobody has posted an accurate solution it I will give a hint on Tuesday.
I will post the solution on Wednesday or Thursday.

chowmein69 · Oct 07, 2007, 07:40 PM // 19:40

27 ways i tihnk .. 9*3

Yellow_lid · Oct 07, 2007, 07:42 PM // 19:42

It's a little more complicated than that.

chowmein69 · Oct 07, 2007, 07:44 PM // 19:44

hmm ... so is it 9 ^ 3 ??
729

Yellow_lid · Oct 07, 2007, 08:40 PM // 20:40

Again, still more complicated, look at the solution for the previous problem. You should try writing them all out in a smaller case, come up with a generalization, and go for the bigger case and see if it follows through.

Aera · Oct 07, 2007, 08:52 PM // 20:52

something like 3x3x3x3x3x3x3x3x3x9^27 or something?

...i hate you for posting this because i'm probably going to be on it all day

*edit*

i misread, it's going to be even harder now...

RedStar · Oct 07, 2007, 09:44 PM // 21:44

I think it's 3^9 : 19683

Caoimhe · Oct 07, 2007, 09:49 PM // 21:49

Remember, though, that your solutions will include the possibility that at least one child will NOT get a piece of candy, if you go for a straight exponential progression. In other words, for each child to get at least one piece of candy, a straight x^y solution won't be accurate.

RedStar · Oct 07, 2007, 09:55 PM // 21:55

Ok, I think I found the answer :
3^9 - 9^3 = 19683 - 729
= 18954

Yellow_lid · Oct 07, 2007, 10:35 PM // 22:35

Sadly, you are not correct, you are still missing something pretty big... Don't think about easy way to obtain the answer, instead try to find the answer the hard way, and look for something easy about the hard way you solved it... And remember what I said about breaking it down into smaller problems first.

I am constantly checking the forum (Via instant e-mail notification). If someone gets the right answer I will post them as correct immediately.

RedStar · Oct 07, 2007, 10:37 PM // 22:37

T_T it's hard....
I'll try to find the answer later today.

chowmein69 · Oct 07, 2007, 10:38 PM // 22:38

what kind of math is this lol

Yellow_lid · Oct 07, 2007, 10:41 PM // 22:41

This my friend is discrete mathematics. (399) it deals with counting methods (see Q1 or Q2) and semi-finite series. It is a Junior level course offered at Oregon State University.

chowmein69 · Oct 07, 2007, 10:41 PM // 22:41

oo i dont tihnk i will ever find the answer -_-

Yellow_lid · Oct 07, 2007, 10:43 PM // 22:43

The highest level of mathematics you actually need to find the answer is 065. Algebra I. But in order to find the solution in a way that would not require you to manually count them all up again you would have to have gone at least through 111 (algebra III).

RedStar · Oct 07, 2007, 10:45 PM // 22:45

I think the answer it somewhere near 68.

A11Eur0 · Oct 08, 2007, 08:28 AM // 08:28

Straight exponential answer minus 7 for the 3 chances that one won't get any, the three chances that two won't get any, and the 1 chance that none will get any.

Lord Sojar · Oct 08, 2007, 09:41 PM // 21:41

I think I would just make a program to solve it for me, thx.

This would be why I got a C- in discrete math. A- in linear algebra.

Yellow_lid · Oct 08, 2007, 10:36 PM // 22:36

LoL. I will get an A in this class and I got a C in Linear Algebra...
And keep it coming... think about actually counting all the ways instead of coming up with a method to count them... 9 and 3 too big? try 4 and 3, or 5 and 3.

amish lifeguard · Oct 09, 2007, 01:46 AM // 01:46

So I just saw this nice thread while perusing the pages. I'm only in pre-calc but I'll get it soon enough!