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We have 2 Americans and we can choose one of them to receive a medal. This can be done in 2 ways.

Choosing a Canadian to receive a medal:
We have 5 Canadians, and we can choose one of them to receive a medal. This can be done in 5 ways.

Choosing a third skater to receive a medal:
After selecting an American and a Canadian, we are left with 17 skaters (19 - 1 American - 1 Canadian). This can be done in 17 ways.

Thus, we have:2 (American's choice).×5(Canadian's choice)×17(third skater's choice)=170 ways
 

Since the balls are indistinguishable, we can think about this problem as placing 4 identical labels into 8 distinct boxes. The first label can be placed in any of the 8 boxes, the second label can be placed in any of the remaining 7 boxes, and so on.  So  8×7×6×5=1680​.

The square root of a complex number is another complex number whose square is equal to the given complex number. The square of i is −1, but there is no complex number whose square is i. Therefore, i does not have any square roots among the complex numbers.

Roger used 15 small Christmas trees * 14 fairy lights/small Christmas tree = 210 fairy lights on small Christmas trees.

The 6 big Christmas trees each had 14 fairy lights/small Christmas tree + 14 fairy lights = 28 fairy lights.

In total, Roger used 210 fairy lights + 6 * 28 fairy lights = 278 fairy lights.

If Roger used 1/3 of the fairy lights to decorate the small Christmas trees, he had 3 * 278 fairy lights = 834 fairy lights at first.

So the answer is 834

There are 21 solutions to the equation u + v + w + x + y + z = 10 where u, v, w, x, y, and z are nonnegative integers, and x is at most 2 or y is at least 3.

To solve this problem, we can use the following cases:

Case 1: x=0 and y>=3. In this case, the only solution is u+v+w+z=10. There is 1 solution in this case.

Case 2: x=1 and y>=3. In this case, the possible solutions are u+v+w+z=9, u+v+w+z=8, and u+v+w+z=7. There are 3 solutions in this case.

Case 3: x=2 and y>=3. In this case, the possible solutions are u+v+w+z=8, u+v+w+z=7, and u+v+w+z=6. There are 3 solutions in this case.

Case 4: x=0 and y<3. In this case, the possible solutions are u+v+w+x+y+z=10, u+v+w+x+y+z=9, u+v+w+x+y+z=8, u+v+w+x+y+z=7, u+v+w+x+y+z=6, u+v+w+x+y+z=5, u+v+w+x+y+z=4, and u+v+w+x+y+z=3. There are 8 solutions in this case.

The total number of solutions is 1+3+3+8=21

To find BC and BZ in triangle AB, we can use the angle bisector theorem.

The angle bisector theorem states that in a triangle, if a line divides one side into two segments, the ratio of the lengths of those segments is equal to the ratio of the lengths of the other two sides.

In triangle AB, let's label BC as "x" and BZ as "y".

According to the angle bisector theorem, we can set up the following ratios:

AY / CY = AB / BC 16 / 16 = 16 / x

Simplifying the equation: 1 = 16 / x

Cross-multiplying: x = 16

So, BC = 16.

Now, let's find BZ using the same theorem:

AY / CZ = AB / BZ 16 / 16 = 16 / y

Simplifying the equation: 1 = 16 / y

Cross-multiplying: y = 16

So, BZ = 16.

Therefore, BC = 16 and BZ = 16 in triangle AB.

And how many participants?