Question

use generating functions to determine the number of different ways 15 identical stuffed animals can be given to 6 children so that each child receivers at least one but no more than 3 stuffed animals

Answer 1

Do you have any ideas for what generating function to use? I'm still thinking about this myself, but I haven't done anything with these in a while (and it wasn't much). :)

Answer 2

I am thinking:
\( \begin{align*}
\quad G(x) &= \left(0 + x + x^2 + x^3 + 0x^4 + \dots \right)^6 \\
&= \left(x + x^2 + x^3\right)^6
\end{align*} \)

The base of the exponent's terms represent that a child could receive a stuffed animal (either a 0(no) or 1(yes)) with the exponent on each term being how many... and all raised to the exponent that 6 children are present to distribute it to.

Then, our solution is simply the coefficient on \(x^15\), which is the number of stuffed animals we have.

Answer 3

\(x^{15}\), not \(x^15\) :)

Answer 4

you are right. Let x(i) is represented for children, we have 6 children. that means we have x(1) +x(2)+x(3) +x(4) + x(5)+x(6) = 15. and 1<= x(i) <= 3. so that we have (x + x^2 +x^3)^6 = 15 . to get the standard form , we must factor x out to get [x(1+x+X^2)]^6 Multiply the left side to (1-x)^6 both numerator and denominator we have the the equation right for denominator only. the numerator . I stuck there.After simplify, I have x^6/ (1-x)^6 - 6 x^9/ (1-x^)6 +15x^12/(1-x)^6 -20 x^15/(1-x)^6 +...there are 3more but i stop there since we have x^15 to calculate, no need to figure out more. but the answer is not right. Can you help me more

Answer 5

sorry, mistake at (x +x^2 +x^3)^6 is the form we must manipulate, take out = 15 (not = 15. my bad.)

Answer 6

\( \begin{align*}
\displaystyle (x + x^2 + x^3)^6 &= (x(1 + x + x^2))^6 \\
&= x^6 (1 + x + x^2)^6 \\
&= x^6 \frac{(1 + x + x^2)^6 (1 - x)^6}{(1 - x)^6}
\end{align*} \)

Am I following that correctly? I am not entirely sure what we are doing here. :)

Answer 7

yes. you are right, but when expand the numerator, you will get what i have. by the way, how can you put the mathematical symbol like this. It is easy to follow but I do not know how to do

Answer 8

I go to eat something, i will be back as soon as I can. Appreciate.

Answer 9

There is something called LaTeX for (in this case) typesetting math equations easily. There is the Equation editor (first button on left under post) to type out basic things in LaTeX, although I've learned enough to just type it out directly. (the editor lacks some things such as a fraction tool, for example -- its \frac{num}{den})

and, okay. Have a good eat. :P

Answer 10

oh, now that i look again, it does include fractions now. cool. It used to not have it. lol

Answer 11

I know it. I used to use it couple times, but sometimes, it doesn't work. ok , back to my problem

Answer 12

k, well one thing I am unsure about is what you are trying to do with the (1 - x)^6 exactly?

Answer 13

\[\frac{ 1 }{(1-X)^6 }= \Sigma \left(\frac{ K+5 }{ 5} \right)^{?}\]

Answer 14

continue aftter ( ) is x^k

Answer 15

in numerator we break them down, expand first to have the form of [something]/(1-x)^6. to have the x^something until we get x^15. at that time, we can figure out the coefficient of x^15

Answer 16

Hey, I need your help, right? what are we doing now?

Answer 17

\[\sum_{k=0}^{\infty}\left(\begin{matrix}k+5 \\ 5\end{matrix}\right)^{} x ^{k}\]

Answer 18

Sorry, I'm just thinking about it. I am not as familiar with this topic, unfortunately

You are able to bump your question up (the "Bump" button below first post) to see if somebody else may be more experienced here. :)

Answer 19

if you have \[\frac{ 1 }{1-x)^6 }\] you can calculate the coefficient of x^ something you can by the value of k, for example if you want to have coefficient of x^78 so k =78 apply to that formula to get the result

Answer 20

anyway, nice to meet you, appreciate when there is someone is willing to help. no matter what the result is. I am happy. thanks a lot

Answer 21

hmm.. so, if i follow this correctly, we'd like to take each term of the expansion and figure the coefficient in the nonclosed 1/(1 - x)^6 for the exponent that 'completes' the x^15:
\( \displaystyle \frac{x^6}{(1 - x)^6} = x^6 (... + Cx^9 + ....) \)

Answer 22

if you have the left side as you post the right side will be \[\sum_{k=0}^{\infty} \left(\begin{matrix}k+5 \\ 5\end{matrix}\right) x^(k+6)\]

Answer 23

sorry I mean x^(k+6)

Answer 24

and if you find coefficient of x^15 at that time 15=k+6 ---> k=9. replace to formula for k you have 9+5=14 choose 5 ; = (14*13*12*11*10)/5*4*3*2*1 =2002

Answer 25

Yep, and we do this for each term:
\(x^6 - 6x^9 + 15x^{12} - 20x^{15} \)

Answer 26

C(9+5,5) is first coefficient, and then C(6+5,5) for next one, C(3+5,5) and C(5,5)
these multiplied with the terms'coefficients as well

C(14,5) - 6C(11,5) + ..

Answer 27

the literally is right, but the answer after getting it is not right at all. since i stop at x^15 (after that number I still have 3 more. so I do not know what happen then

Answer 28

Hm... what number did you get?

Answer 29

36-2772+840-20 LOL

Answer 30

no for the first number, I have wrong copy

Answer 31

the first number is C9,5) right?

Answer 32

first number should be C(9+5,5) or C(14,5), since our value is C(k+5,5) at k=9.

Answer 33

yes, I am sorry, you are right. My bad

Answer 34

so with C(14,5) - 6C(11,5) + 15C(8,5) - 20 C(5,5); I seem to get 50.

Answer 35

2002-2772+840-20=50

Answer 36

yes. it is the right answer. LOL. thanks a loooooot

Answer 37

You're welcome! :)
I see why you went to multiply that (1-x)^6 on now, that is an interesting manipulation. :P

Answer 38

because of you, you figure out my mistake from C(14,5) I got it now. how ever, I have another problem which I have no idea. even my professor, he couldn't give out the right answer . Can you help me

Answer 39

hmm, I might be going to eat for now. You can post it in a new question and I'll probably check back when I'm done? :)

Answer 40

of course. I do it now.