Question

8 children have 9 candy. In how many ways we can divide candy for children? (one children has to have at least one candy) (candy are all different and children are all different)

Answer 1

Assigning the first \(8\) pieces is just \(8!\). Then to assign the last piece there are \(8\) more choices.
This gives us \(8!\times8\)

Answer 2

Tasks done in sequence are multiplied.

Answer 3

but what if I take, for example, 2 children and 3 candy and, if i write down every possibility i get 6, but using your example it should be 2! x 2

Answer 4

Are you sure? Can I see your possibility list?

Answer 5

Let the candies be r, b, g and a comma separating for each kid...

Answer 6

Oh I see the error now...

Answer 7

It matters which candy the 9th candy is.

Answer 8

Okay, so how about 9C8 to determine ways to pick which candies will be the first 8. Then 8! to assign those candies. Lastly 9C1 to pick 9th candy and 8 ways to assign it.

Answer 9

Then for 3 candy 2 kids. it is \(^3C_2 \times 2! \times ^3C_1\times 2\)

Answer 10

Hmmm I wonder...

Answer 11

maybe it should be something like if we have 2 children and 3 candy, then all possibilities are 2^3, with possibilities that one of the children won't get any candy at all, so we have to also take away these 2 possibilities. it works with smaller numbers, but i don't know what about the rest...

Answer 12

Nope that doesn't seem to work either.

Answer 13

Then you can to figure out how to count how many ways a kid can not get candy.

Answer 14

Okay we could do it like this...

Answer 15

Actually never mind I was thinking something wrong.

Answer 16

8^9 is all possibilities with no candy for somone?

Answer 17

All possibilities... even that a kid gets NO candy.

Answer 18

If there are 4 kids... how many ways do you count?

Answer 19

Here is my new idea.
We chose one of the 9 candy which will be the bonus candy.
We then give that to one of the 8 kids.

Then after that we forget about the ninth candy and assign the remaining 8 candy.

So its: \((9\times 8) \times 8!\)

Answer 20

i've counted for 2 children and 3 candy and 2 children and 4 candy and i've written them all down

Answer 21

can you tell me your numbers?

Answer 22

2 children and 3 candy are 6 possibilities and 2,4 are 14 possibilities

Answer 23

I think I figured out my error

Answer 24

Okay so we start off by lining up the kids and have each one draw a random candy. This is \(^9P_8\)

Answer 25

Have you done 3 children 4 candy?

Answer 26

no, i think it is too difficult to count, i should be less than 81 i think

Answer 27

\[
a-bc\\
b-ac\\
c-ab\\
ab-c\\
ac-b\\
bc-a
\]

Answer 28

\[
a-b-cd\\
a-c-bd\\
a-d-bc\\
b-a-cd\\
b-c-ad\\
b-d-ac\\
\]

Answer 29

I shouldn't have been using permutations I think.

Answer 30

Hmmm we gotta split it up into simple tasks.

Answer 31

Okay here is one idea...

1. Choose 1 kid (8)
2. Have the kid choose 2 candy (9C2)
3. Have the remaining kids chose their candies ((8-1)!)

Answer 32

\[
2\times \binom{3}{2}\times (2-1)! = 2\times 3\times 1 = 6
\]

Answer 33

For 3-4: \[
3\times \binom{4}{2}\times (3-1)! = 3\times 6 \times 2= 36
\]

Answer 34

and what if we take 2-4?

Answer 35

\[
8\times \binom{9}{2}\times (8-1)! = 8 \times 36 \times 5040= 1451520
\]

Answer 36

It doesn't work for 2-4 because the problem with 2-4 is that you have many ways of distributing the candy.

Answer 37

I mean you can do 2-2, 1-3, 3-1

Answer 38

My method isn't generalized to that extent. I only works for distributing n among n-1

Answer 39

ok, i understand now, thank you :)

Answer 40

Ummm I wish we had list of 3-4 to check.

Answer 41

I'm getting 36 possibilities so it wouldn't be impossible to list them out.

Answer 42

a.b.cd
a.c.bd
a.d.bc
b.a.cd
b.d.ac
b.c.ad
c.a.bd
c.b.ad
c.d.ab
d.a.bc
d.b.ac
d.c.ad

That's 12, when the third guy gets the candy.
By symmetry you count for the other three to get \(3\times 12=36\)