Nombre de usuarioQuazars
Preguntas 20
Respuestas 48


(I'm not sure how to edit my post, it's supposed to be 30 instead of 40. Derp)


My reasoning led me to the conclusion that since the last cookie is dependent on the one before it the last one can always be ignored.. to more explain how I mean I'll set up the example with two mathematicians and 10 cookies:


No matter how many cookies x the first one gets(0-10, 11 possibilities) the other one will get 11-x(since all cookies need to be distributed)


0 10
1 9
2 8


4 6
5 5
6 4
7 3
8 2
9 1
0 10


- 11 combinations.


When doing this for three mathematicians I follow the same logic that the last persons cookie count is neglectable.


0 0 10
0 1 9
0 ... ...
0 10 0


These are the combinations for when the first person gets no cookies whatsoever. (still 11)


But when the first person gets a cookie the amount decreases by one. And for each cookie he(or she) gets the amount decreases...


1 0 9
1 1 8
1 ... ...
1 9 0


Continue doing this until the first person gets 10 cookies and the rest none and I ended up with:


1+2+3+4+5+6+7+8+9+10+11. Or \(\sum_{i=1}^{11} i\)


I ended up looking through the case when there are four mathematicians(mostly seeing that depending on the value of cookies the first person had the combination count would be that sum, but 11 would decrease by each cookie the first mathematician had. So I set up the formula you can see in the post. At least that was my reasoning. I don't think N choose K applies here, sorry if I wasn't clear enough in my main post. I also sent you a pm, cheers :)

Quazars 19-dic-2016