To determine how many different pairs of dancers could be selected from six males and four females (one male and one female in each pair), we need to approach the problem step by step.

1. Select a male dancer for each pair:
   There are 6 males available, so for the first pair, there are 6 possible choices.

2. Select a female dancer for each pair:
   Once the male dancer is selected, there are 4 females available, so for the first pair, there are 4 possible choices.

3. Determine the total number of pairs:
   Since we are forming 3 pairs, and each pair consists of one male and one female, we multiply the number of ways to select each pair:

   $\text{Total number of pairs} = 6 \times 4 \times 5 \times 3 \times 4 \times 2$

   However, the order of the pairs does not matter, so we divide by the factorial of the number of pairs (3! = 6) to account for the different possible arrangements of the pairs:

   $\text{Total different pairs} = \frac{(6 \times 4) \times (5 \times 3) \times (4 \times 2)}{3!}$

   Calculating this:

   $\text{Total different pairs} = \frac{(6 \times 4 \times 5 \times 3 \times 4 \times 2)}{6} = \frac{2880}{6} = 480$

Therefore, the total number of different pairs that could be selected is 480.

Would you like more details or have any questions?

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Related Questions:

1. How would the number of pairs change if there were 5 males instead of 6?
2. What if there were 5 females instead of 4?
3. How does the formula change if the number of males and females were the same?
4. How can this problem be generalized for any number of males and females?
5. What would be the number of ways to select pairs if the order within each pair mattered?

Tip: Always account for whether order matters in combinatorial problems, as this affects whether you should divide by a factorial.