You have ten indistinguishable cookies that you’ve baked for four of your friends Jianbei, Sione, Sina and Julia. You want to give away all ten of your cookies to your friends, and for each friend to get at least one cookie. You also know that Sione wants an even number of cookies, so he can share them with a friend. In how many ways can you give away your cookies?
I'm having difficulty going about this. Firstly, I thought of breaking down the problem into three separate parts:
Sione has $2$ cookies $\implies$ split $8$ between $3$
Sione has $4$ cookies $\implies$ split $6$ between $3$
Sione has $6$ cookies $\implies$ split $4$ between $3$
From here, I'm a bit stuck on how to find the total, any hints or advice on formula? I got $80$ first time round, but this seems so wrong.
So using the $3$ separate parts above:
$$\text{Total} = 8C3 + 6C3 + 4C3 = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} + \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} + \frac{4 \cdot 3 \cdot 2}{3 \cdot 2 \cdot 1} = 56 + 20 + 4 = 80$$
However, I think this may be wrong as the cookies are all identical e.g. Jianbei getting $2$ cookies and then Sione getting $2$ cookies is the same as Sione getting $2$ cookies and then Jianbei getting $2$ cookies, so I'm thinking maybe this formula would work: $$\frac{n!}{k!(n-k)!}$$
Best Answer
Your strategy is sound, but your answer is indeed incorrect.
Case 1: Sione receives two cookies.
Let $x_i$, $1 \leq i \leq 4$, be the number of cookies distributed to the $i$th friend. Let $x_4$ denote the number of cookies given to Sione. Since a total of ten cookies are distributed and Sione receives two, \begin{align*} x_1 + x_2 + x_3 + 2 & = 10\\ x_1 + x_2 + x_3 & = 8 \tag{1} \end{align*} Since each friend receives at least one cookie, equation 1 is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of two addition signs in the seven spaces between successive ones in a row of eight ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ For instance, if we place an addition sign in the third and seventh boxes, we obtain $$1 1 1 + 1 1 1 1 + 1$$ which corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 1$. The number of such solutions is the number of ways we can select two of the seven spaces in which to place an addition sign, which is $$\binom{7}{2}$$
A particular solution of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = k$$ in the positive integers corresponds to the placement of $n - 1$ addition signs in the $k - 1$ spaces between successive ones in a row of $k$ ones. Therefore, the number of such solutions is
$$\binom{k - 1}{n - 1}$$ since we must choose which $n - 1$ of those $k - 1$ spaces will receive an addition sign.
Case 2: Sione receives four cookies.
Let the variables be assigned as above. Since Sione receives four cookies, \begin{align*} x_1 + x_2 + x_3 + 4 & = 10\\ x_1 + x_2 + x_3 & = 6 \tag{2} \end{align*} Since each friend receives at least one cookie, equation 2 is an equation in the positive integers with $$\binom{6 - 1}{3 - 1} = \binom{5}{2}$$ solutions.
Case 3: Sione receives six cookies.
Let the variables be assigned as above. Since Sione receives six cookies, \begin{align*} x_1 + x_2 + x_3 + 6 & = 10\\ x_1 + x_2 + x_3 & = 4 \tag{3} \end{align*} Since each friend receives at least one cookie, equation 3 is an equation in the positive integers with $$\binom{4 - 1}{3 - 1} = \binom{3}{2}$$ solutions.
Total: Since these cases are mutually exclusive and exhaustive, you can distribute the cookies in $$\binom{7}{2} + \binom{5}{2} + \binom{3}{2} = 21 + 10 + 3 = 34$$ ways.