At the start of the evening, Jim wanted to photograph some of the children of the three families who had contributed the most to the earlier fund-raising activities. One family had three children, one had four and the third had two.

What is the least number of these children that Jim will have to have in a photograph to ensure

*(a) he has at least one pair of siblings in the photograph?* 	

To solve this problem, we can apply the **pigeonhole principle**, which states that if more items (pigeons) are distributed among fewer containers (pigeonholes) than the number of containers, at least one container will hold more than one item.

Here, the **children** represent the "pigeons" and the **families** represent the "pigeonholes." The goal is to find the minimum number of children Jim must photograph to ensure that at least one pair of siblings (children from the same family) is included in the photograph.

### Step-by-step solution:

- **Family 1:** 3 children
- **Family 2:** 4 children
- **Family 3:** 2 children

If Jim wants to avoid having siblings in the photo, he should pick only one child from each family. In this case, Jim could select one child from each of the three families, giving him 3 children (one from each family), without having any siblings in the photo.

- **Scenario without siblings:** 3 children (one from each family).

However, as soon as Jim adds one more child (bringing the total to 4 children), he must select a second child from one of the families. Therefore, he will have a pair of siblings in the photo.

### Conclusion:

The least number of children Jim must include in the photograph to ensure at least one pair of siblings is:

\[
\boxed{4}
\]

Would you like more details on this explanation?

Here are 5 related questions to extend this topic:
1. What if there were 5 families with varying numbers of children? How would that change the solution?
2. How does the pigeonhole principle apply in other real-life situations?
3. What is the maximum number of children Jim can photograph without siblings if there are more than three families?
4. How would the problem change if each family had the same number of children?
5. What is the general formula to determine the minimum number of selections needed to guarantee a pair of siblings for any number of families?

**Tip:** The pigeonhole principle is a powerful tool for ensuring that some conditions are met when distributing objects into groups!

At the start of the evening, Jim wanted to photograph some of the children of the three families who had contributed the most to the earlier fund-raising activities. One family had three children, one had four, and the third had two. What is the least number of these children that Jim will have to have in a photograph to ensure (a) he has at least one pair of siblings in the photograph?

This problem explores the application of the pigeonhole principle to determine the least number of children Jim needs to photograph to ensure at least one pair of siblings. Using the pigeonhole principle, find the minimum number of children selected from three families with varying numbers of children. Suitable for Grades 8-10.

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