Mean babysitter

Medium

Combinatorics Counting

Problem

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?

Original Problem Link: Click here

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:

(4 10) = (6 10) = 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}, f_{b} = 1 + x_{b}, f_{c} = 1 + x_{c}, 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:

(4 - 1 8 + 4 - 1) = (3 11) = 165

Step 3: Computing the Final Answer

The total number of valid distributions is:

(4 10) \times (3 11) = 210 \times 165 = 34, 650