Number of ways to distribute ice cream to children.

Question

A certain school of 10 children is visiting the local ice-cream factory to see how ice-cream is made. After the demonstration, the factory has 15 scoops of vanilla ice-cream and 2 scoops of chocolate ice-cream to distribute to the kids. How many ways can the ice-cream be distributed if each child must receive at least one scoop of ice-cream?

My solution to this is by partitioning the set of ways by the number of chocolate scoop a child will get. Let $C$ be the set of ways to distribute all the ice cream and let $C_1,C_2 \subset C$, where $C_i$ is the set where a child gets $i$ scoops of chocolate ice cream. $C_1 = \ $${10}\choose{2}$$\cdot$${8}\choose{8}$$\cdot$${10}\choose{7}$$=5,400\ ways$, where we must first distribute the 2 scoops of chocolate ice cream to the 10 children (${10}\choose{2}$), then give those who are not given a scoop of chocolate ice cream a scoop of vanilla ice cream (${8}\choose{8}$), and lastly I distribute the remaining 7 vanilla ice cream to all 10 children (${10}\choose{7}$). In the same manner, $C_2 = $${10}\choose{1}$$\cdot$${9}\choose{9}$$\cdot$${10}\choose{6}$ $=2,100$ ways. Hence $C = C_1 + C_2 = 7,500 \ $ ways.

Is this correct?

Answer 1

There are two options (assuming all scoops must be distributed):

One kid gets both scoops of chocolate You need to choose the lucky one ($10$) options, then give one vanilla to the rest. You are left with 6 vanilla scoops to give the 10 children. It is equivalent to the number of solutions to $x_1+\ldots+x_{10}=6$ with $x_i\geq 0$, which is ${10+6-1 \choose 10}$.

Two kids get chocolate You need to choose the two (${10 \choose 2}$) then give the 8 kids a scope of vanilla. You are left with 7 vanilla to give the 10 children, so the same as above with different numbers.

Total $$10 {15 \choose 10} + {10 \choose 2}{16 \choose 10}$$

Answer 2

Total ways to distribute $15V+2C$ among $10$ kids: $$ \binom{10-1+15}{15}\binom{10-1+2}{2} $$ subtract the number of ways one kid gets nothing: $$ 10\times\binom{9-1+15}{15}\binom{9-1+2}{2} $$ add the number of ways two kids get nothing: $$ \binom{10}{2}\times\binom{8-1+15}{15}\binom{8-1+2}{2} $$ and so on. In total: $$ \binom{10-1+15}{15}\binom{10-1+2}{2}-\sum_{k=1}^{9}(-1)^{k-1}\binom{10-k-1+15}{15}\binom{10-k-1+2}{2} $$

Answer 3

First notice that there are two cases for the distribution of $2$ chocolate scoops. Either $2$ of them go to the same child, or they go to $2$ different child.

$2$ chocolate scoops go to the same child
There are 10 ways of giving 2 chocolate scoops to the same child. Every child must get one scoop of ice-cream. There are 10 ways of doing it. Therefore, we give every child a scoop of vanilla ice-cream except the one having two chocolate scoops. After this we have $6$ scoops of vanilla ice cream remained. We can distribute it randomly to the $10$ students. Using the combination with repetition formula, $$10{n+k-1\choose k}=10{10+6-1\choose10}=30030$$

$2$ chocolate scoops go to the $2$ different child
In this case, there are ${10\choose2}$ way to distribute the chocolate scoops. Then give every student a scoop of ice-cream except the $2$ students with chocolate scoops, We have $7$ scoops of ice-cream remained. Again with the combination with repetition formula, we have $${10\choose2}\cdot{10+7-1\choose10} = 8008$$

The answer is thus $30030 +8008 = 38038$.