Curse

coil · Oct 09, 2007, 09:49 PM // 21:49

assuming AAAB=AABA (order doesnt matter, is that the right terminology?) and all 9 candies dont have to be used (as long as each child has 1): 84

1 way (3 candies) + 3 ways (4 candies) + 6 + 10 + 15 + 21 + 28 (9 candies).

Yellow_lid · Oct 09, 2007, 10:00 PM // 22:00

Good, but the candies are different and they all must be handed out in the end... if they were all the same your answer would be correct. I think...

Ok here is your hint:
You are trying to find the number of ways 9 different candies can be dispersed to 3 kids, in such a way that each child gets at least one. The easiest way to find the answer is to take the complement of the set you are after and subtract it from the overall set of possibilities, which as can be seen in Q1 (see first post in thread for link) is 3^9.
You are after the complement of the set of possibilities where all kids get at least one, which is at least one kid gets nothing...

NEW!!!
Keep trying. The person who gets closest will get a point!
10 platinum to the first person to 3 points!

JoeKnowMo · Oct 09, 2007, 10:07 PM // 22:07

For stuff like this I usually rely on my IQ score.
28.

Yellow_lid · Oct 09, 2007, 10:11 PM // 22:11

Sorry, 28 is not the answer.

Zeuki · Oct 09, 2007, 10:16 PM // 22:16

63.

9 Candies, 7 Possible solutions per piece of candy since 8 and 9 candies per child is not possible due to atleast one child missing one piece of child on each.

JoeKnowMo · Oct 09, 2007, 10:17 PM // 22:17

Quote:

Originally Posted by Yellow_lid

Sorry, 28 is not the answer.

I assumed that each unique arrangement would be counted once. That is to say that I thought 7,1,1 was different from 1,7,1 which was different from 1,1,7.

Since I misunderstood the question, my IQ is now 12.
So that's my answer: 12.

Please don't lower my IQ further.

Yellow_lid · Oct 09, 2007, 10:17 PM // 22:17

but the first candy could go to any of the three children and those would all be different cases wouldn't they?

Yellow_lid · Oct 09, 2007, 10:18 PM // 22:18

THE CANDIES ARE DIFFERENT...
CANDY 1 GOING TO CHILD A and the rest going to B is NOT the same as candy 1 going to B and the rest going to A.

Zeuki · Oct 09, 2007, 10:20 PM // 22:20

My apologies, I forgot a step.

9 Candies, 7 Solutions per candy, 63 per child, 189 solutions total.

Yellow_lid · Oct 09, 2007, 10:24 PM // 22:24

Zeuki, Look at the solution for Q1 and try looking at it the other way. You are not trying to find out how many solutions there are for children, you are looking for the number of solutions per candy...
As stated in Q1. The first candy could go to any of the three children. 3 possibilities. The second candy can go to any of the three children for each of the first three options.
AA AB AC
BA BB BC
CA CB CC
These are all different possibilities, and they are only for 2 candies!

JoeKnowMo · Oct 09, 2007, 11:44 PM // 23:44

Alright. Now that I finally understood the question (I think).

I tried following your advice and working with smaller numbers first.

Extrapolating from those I've got: 3 ^ 6 = 729
729 * 9 = 6561

So it seems like there are 3 ^ 8 possibilities, which is 6,561.

Mercury Angel · Oct 10, 2007, 12:18 AM // 00:18

Ok, if I understand this right, there are 6 basic patterns;
711, 621, 531, 522, 441, 432

For 711 distributions, that should be 72 possibilities.
9*8 possibilities of distributing the 9 candies such that 2 children each get one piece, and the third child gets the rest. There are three children, so there are 3 different children to get 7 pieces. The total possible 711 distributions with variations should be 216.
For 621, there are 140 possible patterns where a specific child receives 6, which should mean 420 variations for all three children.
(7^2 + 6^2 + ... 1^2)

For 531, that should mean 336 possible patterns where a specific child receives 5 pieces of candy, and then 1008 for all three children.
(7^2 + 2* 6^2 + ... 7 * 1^2)

Am I on the right track here for solving it the long way?

Alternatively, if you distribute 9 pieces of candy among 2 children (2^9), a specific child with none, there are 512 possibilities. This includes 900 and 090 I believe. Any other solutions can be arranged such that there are 6 possibilities (Ex: 810, 801, 108, 180, 018, 081), so I subtract 2, and that leaves me with 510, and multiply it by 6 for 3060. Then, I tack on the 900, 090, and 009 solutions, and end up with 3063.

Therefore, I'm guessing there are 16620 possible solution sets wherein each child receives at least one piece of candy from 9 unique pieces of candy?

I Might Avenge U · Oct 10, 2007, 12:19 AM // 00:19

1 1/2 midgets!

coil · Oct 10, 2007, 12:31 AM // 00:31

Quote:

Originally Posted by Yellow_lid

Good, but the candies are different and they all must be handed out in the end... if they were all the same your answer would be correct. I think...

yea..assumed they werent all unique. so...+.5 pnts ?

otherwise, if every small step is considered unique... i'm not a math guy (anymore) but: 3^9-3(3-1)^9+3(3-2)^9 = 18150

?

Yellow_lid · Oct 10, 2007, 01:23 AM // 01:23

Quote:

Originally Posted by coil

otherwise, if every small step is considered unique... i'm not a math guy (anymore) but: 3^9-3(3-1)^9+3(3-2)^9 = 18150?

Congratulations!!!
This is not only the correct numerical answer but it was obtained correctly too!!!

Score: Coil=1

I will post a more detailed version of the answer after dinner.

Yellow_lid · Oct 10, 2007, 01:49 AM // 01:49

Q2 In how many ways can you pass out candy if each child must receive at least one piece?
From Q1 It was shown that there are 3^9 different ways to hand out the candies to three children in general, this is the list of functions from that maps candies to children.
Now we must count how many of those functions are onto, where every child gets at least one candy. This can be viewed as the entire set of possible functions minus the complement of the set where each child gets at least one, which is at least one child gets nothing.
How many possibly functions are there where at least one child gets nothing? In order for that to happen, all the candies must go to the same two children. Similar to what was done in Q1.
There are two possibilities for the first candy (A or B), it must not go to child C.
2
There are two possibilities for the second candy (A or B), for each of the possibilities for the first candy. It also cannot go to child C.
2∙2
There are two possibilities for the third candy (A or B), for each combination of the first and second. Again, it cannot go to child C.
2∙2∙2
Follow the logical pattern, and you will get
2∙2∙2∙2∙2∙2∙2∙2∙2 = 2^9
There are 2^9 possibilities where children A and B get everything.
There are 2^9 possibilities where children A and C get everything.
There are 2^9 possibilities where children B and C get everything.
So far we see that there are 3∙2^9 possibilities where at least one child gets nothing. The possibility that one child gets everything was also included in the 3∙2^9. In fact, it was included twice. The double counted possibilities must be removed, and we counted each of the three possibilities where one child gets everything twice. We counted A and B get everything, or just A, or just B. We counted A and C get everything, or just A (again), or just C. Finally we counted B and C get everything, or just B (again), or just C (Again).
As you can see we counted A, B, and C twice, so we must subtract three from our complementary set of solutions. The complement Set now looks like:

(3∙2^9-3)

Originally you can note that we were taking the whole set minus the complement of the set that we actually want. The final equation looks like this.

3^9-3∙2^9+3 = 18,150
There are 18,150 different ways to hand out nine candies to three children where each child will get at least one candy

Furthermore it can be shown that the formula for any set of candies and kids where each kid must have one candy is...
K=kids, C=Candy
PK^C-P(K-1)^C+P(K-2)^C-...+P(K-K)^C
There P are the pascal's coefficient on the K+1st tier of the triangle.

Yellow_lid · Oct 10, 2007, 01:49 AM // 01:49

Look for the next Question on Friday!

freaky naughty · Oct 12, 2007, 01:43 AM // 01:43

Holy sh*t lol I'm in freshman year of high school, so I guess i'll go farm that 10 plat instead of answering these.

Yellow_lid · Oct 12, 2007, 01:55 AM // 01:55

Quote:

Originally Posted by freaky naughty

Holy sh*t lol I'm in freshman year of high school, so I guess i'll go farm that 10 plat instead of answering these.

It's not a here's how you can earn 10K it's more of a just for fun thing, with a prize for best answer... lol

Yellow_lid · Oct 15, 2007, 10:38 PM // 22:38

My apologies to everyone who has been waiting for the next in the series of the math questions thread. I have not encountered a problem of quite the right caliper for this thread series... but rest assured, I have another homework assignment due Thursday. and i will try to get one from there by then.