Friends and Chicken ( re-revised )

Firstly count the number of ways that the four friends could order eleven pieces of chicken, if it does not matter if any of the friends does not order anything. This is equal to the number of ways of putting $11$ objects into $16$ boxes, which is equal to $\binom{26}{11}$ by a standard stars-and-bars argument.

Let $X_1$ be the number of ways of buying eleven pieces of chicken if one specific friend does not get anything. This is simply the number of ways of distributing the $11$ pieces of chicken amongst the other three friends, which is the number of ways of putting $11$ objects into $12$ boxes, which is $\binom{22}{11}$ .

Let $X_2$ be the number of ways of buying eleven pieces of chicken if two specific friends do not get anything. This is simply the number of ways of distributing the $11$ pieces of chicken amongst the other two friends, which is the number of ways of putting $11$ objects into $8$ boxes, which is $\binom{18}{11}$ .

Let $X_3$ be the number of ways of buying eleven pieces of chicken if three specific friends do not get anything. This is simply the number of ways of the fourth friend buying all the chicken, which is the number of ways of putting $11$ objects into $4$ boxes, which is $\binom{14}{11}$ .

It is not possible for all four friends not to buy chicken.

Using the Inclusion-Exclusion Principle, the number of ways of buying the chicken pieces, with at least one friend not getting any chicken, is $4X_1 - 6X_2 + 4X_3$ and hence the number of ways pf buying the chicken pieces, with every friend getting at least one piece of chicken is $\binom{26}{11} -4\binom{22}{11} + 6\binom{18}{11} - 4\binom{14}{11} = \boxed{5093920}$

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More generally, let $N(a,b,c)$ be the number of ways in which $a$ friends can by a total of $b \ge a$ pieces of chicken, which can can come in $c$ different types, with each friend getting at least one piece of chicken.

Let $S$ be the collection of $a$ -tuples $(u_1,u_2,...,u_a)$ of positive integers which add to $b$ . Then $S$ is the set of possible distributions of total number of chicken pieces amongst the $a$ friends. For each such distribution, the first friend has to arrange his $u_1$ pieces of chicken into the $c$ different types that are available, and this can be done in $\binom{u_1+c-1}{c-1}$ ways. The other friends have to do the same. Thus $N(a,b,c) \; = \; \sum_{(u_1,u_2,...,u_a) \in S} \binom{u_1+c-1}{c-1}\binom{u_2+c-1}{c-1} \cdots \binom{u_a+c-1}{c-1}$ and this is the coefficient of $X^b$ in the power series expansion of $\left(\sum_{u \ge 1}\binom{u+c-1}{c-1}X^u\right)^a$ Now $\sum_{u \ge 1}\binom{u+c-1}{c-1}X^u \; = \; \frac{1}{(c-1)!}\frac{d^{c-1}}{du^{c-1}} \sum_{u=1}^\infty X^{u+c-1} \; = \; \frac{1}{(c-1)!} \frac{d^{c-1}}{du^{c-1}} \frac{X^c}{1-X} \; = \; \frac{1}{(1-X)^c} - 1$ and so $N(a,b,c)$ is the coefficient of $X^b$ in the power series expansion of $\left(\frac{1}{(1-X)^c} - 1\right)^a$ Again, we obtain $N(4,11,4) = 5093920$ .

The sledge hammer approach:

Enumerate the types of sums of $11$ with $4$ summands ( the number of ways for 4 friends to order 11 pieces of chicken in total ) beginning with the largest summand $8$ and work your way down. The various sums and the number of ways to order them is shown below:

$\displaystyle \begin{matrix} & \text{Sums} \quad \quad \quad & \text{Combinations} \\ & 8 + 1 + 1 + 1 &{4\choose1} = N_{8111} =4 \\ & 7 + 2 + 1 + 1 & {4\choose 2} \cdot 2 = N_{7211} =12 \\ & 6 + 3 + 1 + 1 & {4\choose 2} \cdot 2 = N_{6311} = 12 \\ & 6 + 2 + 2 + 1 & {4\choose 2} \cdot 2 = N_{6221} = 12 \\ & 5 + 2 + 2 + 2 & {4\choose 1} = N_{5222} = 4 \\ & 5 + 3 + 2 + 1 & 4! = N_{5321} = 24 \\ & 5 + 4 + 1 + 1 & {4\choose 2} \cdot 2 = N_{5411} = 12 \\ & 4 + 4 + 2 + 1 & {4\choose 2} \cdot 2 = N_{4421} = 12 \\ & 4 + 3 + 2 + 2 & {4\choose 2} \cdot 2 = N_{4322} = 12 \\ & 4 + 3 + 3 + 1 & {4\choose 2} \cdot 2 = N_{4331} = 12 \\ & 3 + 3 + 3 + 2 & {4\choose 1} = N_{3332} = 4 \end{matrix}$

Now that we have the ways in which the friends can order the number of pieces, let us move on the number of ways each person can order chicken pieces to fill said order. The friends do this by choosing a type, or multiple types of chicken pieces (wings, legs, breasts, or thighs) in amounts that add up to their individual order.

For a person that gets $1$ piece of chicken $S_1$ they can do this in $4$ ways ( the number of ways of choosing $1$ type of $4$ :

$S_1 = {4 \choose 1} = 4$

For a person that gets $2$ pieces of chicken $S_2$ they can do this in $10$ ways: They can fill the order with all $1$ type or they may select $2$ out of the $4$ types ( $1$ type1 $+$ $1$ type2 ):

$\begin{matrix} \displaystyle S_2 = &\displaystyle {4 \choose 1} \quad + &\displaystyle {4 \choose 2} \cdot {1 \choose 1} = 10 \\ \quad \\ & BB,LL, &WB,WL,WT \\ & TT,WW & LB,LT,TB \end{matrix}$

For a person that gets $3$ pieces of chicken $S_3$ they can do this in $20$ ways: They can fill the order with all $1$ type , they may select $2$ out of the $4$ types ( $6$ ways ) in $2$ ways each ( $1$ type1 $+$ $2$ type2 or $2$ type1 $+$ $1$ type2 ), or they may select $3$ out of the $4$ types ( $4$ ways ) in $1$ way ( ( $1$ type1 $+$ $1$ type2 + $1$ type3 ):

$\displaystyle S_3 = \displaystyle {4 \choose 1} \quad + \displaystyle {4 \choose 2} \cdot {2 \choose 1} + \displaystyle {4 \choose 3} \cdot {2 \choose 2} = 20$

We keep on with this pattern:

$\displaystyle S_4 = \displaystyle {4 \choose 1} + \displaystyle {4 \choose 2} \cdot {3 \choose 1} + \displaystyle {4 \choose 3} \cdot {3 \choose 2} + \displaystyle {4 \choose 4} \cdot {3 \choose 3} = 35$

Now that we’ve made it to all 4 types as a selection we can write a general way to order $i$ pieces in up to $4$ types:

$\Large S_i = \Large \sum_{j=0}^3 \left[ {4 \choose j+1} \cdot {i -1 \choose j } \right]$

Thus, for $4$ friends $a,b,c,d$ the number of ways to order $w+x+y+z$ pieces is given by:

$\Large \sum \left( N_{wxyz} \cdot S_w \cdot S_x \cdot S_y \cdot S_z \right)$

Example: $\left( w = 8, x = 1, y = 1, z =1 \right)$

$N_{8111} \cdot S_8 \cdot {S_1}^3 = 4 \cdot \left[ \displaystyle {4 \choose 1} + \displaystyle {4 \choose 2} \cdot {7 \choose 1 } + \displaystyle {4 \choose 3} \cdot {7 \choose 2 } + \displaystyle {4 \choose 4} \cdot {7 \choose 3} \right] \cdot \displaystyle {4 \choose 1}^3 = 42,240$

Repeating this calculation for each of the enumerated sums above, the total number from the summation is: $\Large 5,093,920 \, \text{ways}$