Basketball Team Combinations: Solving Selection Scenarios

Hey guys! Let's dive into a super interesting problem about forming a basketball team. We have 40 students, and we need to pick 5 of them to represent IPB. The question is, how many different teams can we create under various conditions? We'll break down this problem step by step, making sure everyone understands the logic and math behind it. So, grab your thinking caps, and let’s get started!

(a) No Restrictions: Calculating $C(40, 5)$

When we talk about combinations with no restrictions, things are pretty straightforward. Basically, we want to know how many ways we can choose 5 players out of 40, without worrying about specific positions or roles. This is a classic combination problem, and we use the formula for combinations, which is often written as $C(n, k)$ or sometimes as “n choose k.” In our case, n is 40 (the total number of students), and k is 5 (the number of players we want to select). So, we’re looking for $C(40, 5)$.

The formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

Where "!" denotes the factorial. The factorial of a number is the product of all positive integers up to that number. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Plugging in our values, we get:

$C(40, 5) = \frac{40!}{5!(40-5)!} = \frac{40!}{5!35!}$

Now, let’s break this down. 40! is a massive number, but we don't need to calculate it fully. We can simplify the expression by canceling out terms. Remember, 40! = 40 × 39 × 38 × ... × 36 × 35!. So, we can rewrite the expression as:

$C(40, 5) = \frac{40 × 39 × 38 × 37 × 36 × 35!}{5! × 35!}$

The 35! in the numerator and denominator cancel each other out, which makes our calculation much easier:

$C(40, 5) = \frac{40 × 39 × 38 × 37 × 36}{5!}$

Now we need to calculate 5!. We know that 5! = 5 × 4 × 3 × 2 × 1 = 120. So, we have:

$C(40, 5) = \frac{40 × 39 × 38 × 37 × 36}{120}$

We can simplify further by canceling out common factors. For example, 40 divided by 5 is 8, and 36 divided by 4 is 9. Continuing the simplification:

$C(40, 5) = \frac{8 × 39 × 38 × 37 × 9}{3 × 2 × 1}$

$C(40, 5) = 8 × 13 × 19 × 37 × 3$

Now, let's multiply these numbers together:

$C(40, 5) = 658,008$

So, there are 658,008 different ways to form a basketball team of 5 players from 40 students if there are no restrictions. Wow, that's a lot of teams! This illustrates how quickly combinations can grow as the numbers get larger. Understanding this basic combination calculation is super important for tackling more complex scenarios. We’ve nailed the first part – now let’s move on to the second, which involves some specific roles within the team.

(b) Dividing the Team: 2 Back Players and 3 Others

Okay, guys, this part gets a little more interesting! Now we’re not just picking any 5 players; we need to form a team with 2 back players and 3 other players. This adds a layer of complexity because we have to consider the specific positions. We'll use the same combination formula, but we'll apply it in stages. First, we’ll choose the back players, and then we’ll choose the remaining players. The key here is to multiply the number of ways to choose each group to get the total number of team combinations.

Let’s assume that all 40 students are eligible for any position. So, we need to figure out how many ways we can choose 2 back players from 40 students. This is another combination problem, $C(40, 2)$.

Using the combination formula:

$C(40, 2) = \frac{40!}{2!(40-2)!} = \frac{40!}{2!38!}$

Again, we can simplify:

$C(40, 2) = \frac{40 × 39 × 38!}{2! × 38!} = \frac{40 × 39}{2 × 1}$

$C(40, 2) = \frac{40 × 39}{2} = 20 × 39 = 780$

So, there are 780 ways to choose 2 back players from the 40 students. Now, we need to choose the remaining 3 players. Since we’ve already chosen 2 players, we have 38 students left to choose from. We want to pick 3 players, so we need to calculate $C(38, 3)$.

Using the combination formula:

$C(38, 3) = \frac{38!}{3!(38-3)!} = \frac{38!}{3!35!}$

Simplifying:

$C(38, 3) = \frac{38 × 37 × 36 × 35!}{3! × 35!} = \frac{38 × 37 × 36}{3 × 2 × 1}$

$C(38, 3) = \frac{38 × 37 × 36}{6} = 38 × 37 × 6 = 8436$

There are 8,436 ways to choose the remaining 3 players from the remaining 38 students. Now, here’s the crucial part: to find the total number of ways to form the team with 2 back players and 3 other players, we multiply the number of ways to choose the back players by the number of ways to choose the other players:

Total combinations = $C(40, 2) × C(38, 3) = 780 × 8436$

Total combinations = 6,580,080

So, there are a whopping 6,580,080 different ways to form the basketball team if it must consist of 2 back players and 3 other players. Notice how adding this condition significantly increased the number of possible teams. This is because we're now considering specific roles, which multiplies the possibilities.

Key Takeaways

• When there are no restrictions, we use the basic combination formula $C(n, k)$ to find the number of ways to choose k items from n items.
• When there are specific conditions (like requiring certain positions), we break the problem into stages, calculate the combinations for each stage, and then multiply the results.

Conclusion: Mastering Combinations

Alright, guys, we’ve tackled a challenging problem and broken it down into manageable parts. We started with a simple combination calculation and then moved on to a more complex scenario involving specific team roles. Understanding how to approach these types of problems is super useful, not just in math class but also in real-life situations where you need to figure out the number of possibilities.

The key takeaway here is that combinations are powerful tools for counting possibilities, and knowing how to apply them can help you solve a wide range of problems. Whether you're forming a basketball team, planning an event, or even just trying to figure out how many different outfits you can make from your wardrobe, combinations are your friend. Keep practicing, and you’ll become a pro at these in no time!

Remember, the main formulas we used are:

• Combination formula: $C(n, k) = \frac{n!}{k!(n-k)!}$
• When combining multiple selections, we multiply the combinations of each selection.

I hope this explanation was clear and helpful! If you have any more questions or want to explore other types of counting problems, let me know. Keep up the great work, guys, and happy calculating!