Suppose there are 3 distinct green balls, 4 distinct red balls.and 5 distinct blue balls. The balls are to be grouped into pairssuch that the balls in any pair have different colors. In how manydifferent ways can such six pairs be formed?

To solve this problem, let's break down the possible pairings based on the requirement that each pair must contain balls of different colors.

### Step 1: Understanding the Pairings
Given:
- **3 green balls (G1, G2, G3)**
- **4 red balls (R1, R2, R3, R4)**
- **5 blue balls (B1, B2, B3, B4, B5)**

We need to form **6 pairs** such that each pair consists of balls of **different colors**.

### Step 2: Analyzing Possible Pair Combinations
Since each pair must have different colors, the valid pairs can be:
- **Green-Red (G-R)**
- **Green-Blue (G-B)**
- **Red-Blue (R-B)**

Since there are 3 green balls, 4 red balls, and 5 blue balls, we’ll need to carefully assign each ball into one of these pair types to ensure we use all balls.

### Step 3: Determining Pair Counts
To make six pairs in total, we can break it down as follows:
1. **Green-Red pairs**: 3 pairs (since we have exactly 3 green balls, they can each be paired with a red ball).
2. **Green-Blue pairs**: 0 pairs (since we've used all green balls in Green-Red pairs).
3. **Red-Blue pairs**: The remaining 3 pairs (since we still have red and blue balls available).

### Step 4: Calculating Choices for Each Type of Pair
#### Green-Red Pairs:
- We select one of the **4 red balls** to pair with each of the **3 green balls**.
- There are \( \binom{4}{3} = 4 \) ways to select 3 red balls from the 4.
- For each selection of 3 red balls, they can be paired with the 3 green balls in \( 3! = 6 \) ways.

Thus, the total number of ways to create the Green-Red pairs is:
\[
4 \times 6 = 24
\]

#### Red-Blue Pairs:
- We have **1 remaining red ball** and **5 blue balls**. We need to form **3 Red-Blue pairs** from these.
- We select 3 of the 5 blue balls to pair with the 3 remaining red balls. This can be done in \( \binom{5}{3} = 10 \) ways.
- For each selection, the 3 chosen blue balls can be arranged with the 3 red balls in \( 3! = 6 \) ways.

Thus, the total number of ways to create the Red-Blue pairs is:
\[
10 \times 6 = 60
\]

### Step 5: Calculating the Total Number of Pairings
Now, we multiply the number of ways to form the Green-Red pairs and the Red-Blue pairs:
\[
24 \times 60 = 1440
\]

### Final Answer
The total number of ways to form the six pairs is:
\[
\boxed{1440}
\]

Would you like a deeper explanation on any part of this solution, or have any follow-up questions?

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Here are some questions related to this problem:
1. What if there were 4 distinct green balls instead of 3?
2. How would the solution change if all balls of each color were indistinguishable?
3. Can we use a combinatorial formula to generalize this for any number of distinct balls?
4. How would the problem change if each pair could be of the same color?
5. What if we had to ensure each color was paired with each other color at least once?

**Tip:** When solving combinatorial problems, organizing possible outcomes step-by-step can simplify complex counting tasks.

Suppose there are 3 distinct green balls, 4 distinct red balls, and 5 distinct blue balls. The balls are to be grouped into pairs such that the balls in any pair have different colors. In how many different ways can such six pairs be formed?

Explore how to solve a combinatorial problem involving grouping distinct green, red, and blue balls into six pairs with different colors. This solution covers counting principles, including combinations and permutations, suitable for Grades 10-12.

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