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Team A and Team B are competing against each other in a game of tug-of-war. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups?
Let's start by understanding exactly what we're being asked to find. Team A has 3 males and 3 females, and they want to line up in a very specific pattern: male, female, male, female, male, female.
Think of it like this: imagine 6 chairs in a row, numbered 1 through 6. The team has already decided that:
The question is asking: "In how many different ways can they fill these 6 chairs while following this male-female-male-female-male-female pattern?"
Process Skill: TRANSLATE - Converting the problem language into a clear mathematical setup
Now let's think about what this constraint means. Since the pattern is already fixed, we're not choosing which positions will be male or female - that's already decided for us.
What we ARE choosing is:
This is like having two separate, independent problems:
Let's count the arrangements for each group separately.
For the males (positions 1, 3, 5):
Let's call the males M1, M2, and M3. For position 1, we can choose any of the 3 males. Once we've chosen someone for position 1, we have 2 remaining males to choose from for position 3. Finally, for position 5, we have only 1 male left.
So the number of ways to arrange the males = \(3 \times 2 \times 1 = 6\) ways
For the females (positions 2, 4, 6):
Similarly, let's call the females F1, F2, and F3. For position 2, we can choose any of the 3 females. For position 4, we have 2 remaining females to choose from. For position 6, we have 1 female left.
So the number of ways to arrange the females = \(3 \times 2 \times 1 = 6\) ways
Here's the key insight: the male arrangements and female arrangements happen completely independently of each other. Whatever way we arrange the males has no effect on how we can arrange the females, and vice versa.
When we have independent events, we multiply the number of possibilities.
Think of it this way: for each of the 6 ways to arrange the males, we can pair it with any of the 6 ways to arrange the females.
Total number of lineups = (Ways to arrange males) × (Ways to arrange females)
Total number of lineups = \(6 \times 6 = 36\)
The answer is 36 different possible lineups, which corresponds to choice D.
To verify: We have 6 ways to fill the male positions (1, 3, 5) and 6 ways to fill the female positions (2, 4, 6). Since these choices are independent, we multiply: \(6 \times 6 = 36\) total lineups.
Students often think they need to count how many ways to arrange 6 people in male-female-male-female-male-female pattern, including choosing which positions are male vs female. They fail to recognize that the alternating pattern is already fixed and only the specific assignment of individuals to their designated positions needs to be counted.
Students may attempt to solve this as a straightforward permutation of 6 people (\(6!\)) and then try to apply some constraint factor, missing that this is actually two independent permutation problems (3 males in male positions, 3 females in female positions) that need to be multiplied together.
Some students may think about choosing 3 positions out of 6 for males (which would involve combinations), not realizing that the positions are predetermined by the alternating pattern requirement.
Students may incorrectly calculate the number of ways to arrange 3 males in 3 positions or 3 females in 3 positions. For example, they might calculate \(3^2 = 9\) instead of \(3! = 6\), or use some other incorrect formula.
Even when students correctly identify that there are 6 ways to arrange males and 6 ways to arrange females, they may add these numbers (\(6 + 6 = 12\)) instead of multiplying them (\(6 \times 6 = 36\)), failing to apply the fundamental counting principle for independent events.
No likely faltering points - once students correctly calculate 36, this directly matches answer choice D with no additional interpretation needed.