To solve this problem, we can break it down into two parts: awarding medals to the boys and awarding medals to the girls.

Boys:

There are 4 boys, and we need to award three medals (gold, silver, and bronze). This is a permutation problem because the order of the medals matters. The number of ways to choose and arrange 3 boys from 4 is given by:

$P(4, 3) = \frac{4!}{(4 - 3)!} = 4 \times 3 \times 2 = 24$

Girls:

There are 13 girls, and we need to award three medals to them as well. Again, this is a permutation problem, and the number of ways to choose and arrange 3 girls from 13 is given by:

$P(13, 3) = \frac{13!}{(13 - 3)!} = 13 \times 12 \times 11 = 1716$

Total number of ways to award the six medals:

Since the selection of boys and girls are independent, we multiply the two results together to find the total number of possible ways to award the six medals:

$\text{Total Ways} = 24 \times 1716 = 41,184$

Thus, there are 41,184 different ways the six medals can be awarded.

Would you like more details on this solution or have any further questions?

Here are 5 related questions you might find interesting:

1. How would the answer change if 5 boys were competing instead of 4?
2. What if there were no distinction between the medals (just 3 winners for boys and 3 for girls)?
3. How would the problem change if the order of awarding medals didn’t matter?
4. How many ways can you assign just the gold medal to both boys and girls?
5. What is the probability of a specific boy and girl winning gold medals?

Tip: When the order matters in selecting items, it is a permutation problem; if the order doesn’t matter, it's a combination.