Problem Statement
10 kids are really hungry! Their babysitter has 12 units of food to give. However, she decides she only wants to give 4 of the children food. How many ways can she distribute the food units such that 6 of the children are hungry (receive no food), and the other 4 children receive at least 1 unit of food each?
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Solution
Step 1: Understanding the Required Tasks
To distribute the food, we need to:
- Choose 4 children who will receive food (equivalently, choose 6 children who won’t).
- Distribute 12 units of food among these 4 children, ensuring that each child gets at least 1 unit.
Step 2: Calculating the Number of Ways
Step 1: Choosing 4 Children
Since we need to select 4 children out of 10, the number of ways to do this is:
\[\binom{10}{4} = \binom{10}{6} = 210\](since choosing 4 to receive food is the same as choosing 6 to not receive food).
Step 2: Distributing the Food
Let the food received by the four chosen children be represented as:
\[f_{a} + f_{b} + f_{c} + f_{d} = 12\]where \(f_{a}, f_{b}, f_{c}, f_{d}\) represent the food units received by each of the 4 children.
Since each child must get at least 1 unit, we make the substitution:
\[f_{a} = 1 + x_{a}, \quad f_{b} = 1 + x_{b}, \quad f_{c} = 1 + x_{c}, \quad f_{d} = 1 + x_{d}\]Rewriting the equation:
\[(1 + x_{a}) + (1 + x_{b}) + (1 + x_{c}) + (1 + x_{d}) = 12\]which simplifies to:
\[x_{a} + x_{b} + x_{c} + x_{d} = 8\]This is now a classic “stars and bars” problem, where we distribute 8 extra food units among 4 children. The number of ways to do this is:
\[\binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = 165\]Step 3: Computing the Final Answer
The total number of valid distributions is:
\[\binom{10}{4} \times \binom{11}{3} = 210 \times 165 = 34,650\]Therefore, the total number of ways is \(\mathbf{34650}\)
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