Question

In how many different ways can eight cookies be distributed among three distinct children, if each child receives at least two cookies and no more than four cookies

Answer 1

I think there can be 6 different ways . They are as follows :

Child 1 Child 2 Child 3
4 cookies 2 cookies 2 cookies
2 cookies 4 cookies. 2 cookies
2 cookies. 2 cookies. 4 cookies
3 cookies. 3 cookies. 2 cookies
3 cookies. 2 cookies. 3 cookies
2 cookies. 3 cookies. 3 cookies
This is a question of combination and I don’t know the formula , so this is the only way I could solve it, I hope it helps.

Answer 2

Answer:

Step-by-step explanation:

Using Generating Function we can solve this.

in general we can write

2,3,4 are possible division of cookies.

thus,

$(x^{2} +x^{3} +x^{4})$

consider that there are 3 children so

$(x^{2} +x^{3} +x^{4}) +(x^{2} +x^{3} +x^{4}) +(x^{2} +x^{3} +x^{4})$ $= (x^{2} +x^{3} +x^{4}) ^{3}$

direct way to find out with    $a{k} x^{k}$

$(1+x+x^{2} +x^{2} +.........+x^{2} )=\frac{(1-x^{(n+1)}) }{(1-x)}$

we can apply it here:

$(x^{2} +x^{3} +x^{4}) ^{3}= x^{6}(1+x+x^{2})^{3}$

                        $=x^{6}(\frac{1-x^{3} }{1-x})^{3}$

                        $=x^{6}(1-x^{3}) ^{3} (1-x)^{-3}$

                        $= x^{6} (1-3x^{3}+3x^{6}-x^{9}) +$$\left \{∑({{3+i-1} \atop {i}} )\right x^{i}$

We applied the Binomial Theorem for a negative exponent in the last step.

                         $\left \{( {{3+2-1} \atop {2}}) \right.=6$