We are tasked with forming a committee of 6 U.S. senators that consists of 3 Democrats and 3 Republicans. There are 69 Democratic senators and 31 Republican senators to choose from.

Step 1: Choosing 3 Democrats

From the 69 Democratic senators, we need to select 3. The number of ways to choose 3 Democrats from 69 can be calculated using the combination formula:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

So, the number of ways to choose 3 Democrats from 69 is:

$\binom{69}{3} = \frac{69 \times 68 \times 67}{3 \times 2 \times 1} = 52394$

Step 2: Choosing 3 Republicans

Similarly, we need to choose 3 Republicans from the 31 Republican senators. The number of ways to choose 3 Republicans from 31 is:

$\binom{31}{3} = \frac{31 \times 30 \times 29}{3 \times 2 \times 1} = 4495$

Step 3: Calculating the Total Number of Ways

To form the committee, we need to select both 3 Democrats and 3 Republicans. Since the selections are independent, the total number of ways to form the committee is the product of the two combinations:

$\text{Total number of ways} = \binom{69}{3} \times \binom{31}{3} = 52394 \times 4495 = 235,872,730$

Thus, the total number of ways to form the committee is:

$\boxed{235,872,730}$

Would you like more details, or do you have any questions?

Here are 5 related questions to expand this topic:

1. How would the result change if we needed to select 4 Democrats and 2 Republicans instead?
2. What if there were only 60 Democratic senators? How would that affect the outcome?
3. Can you find the probability that a randomly chosen committee has exactly 3 Democrats and 3 Republicans?
4. How would the problem change if each party had the same number of senators (say, 50 each)?
5. In how many ways can we form a committee of 6 senators with at least 2 Democrats?

Tip: Remember, combinations count selections where order doesn't matter. This is why we use $\binom{n}{r}$ instead of permutations.